If $\overrightarrow{a}$ and $\overrightarrow{b}$ are two non-zero perpendicular vectors, then a vector $\overrightarrow{y}$ satisfying equations $\overrightarrow{a}.\overrightarrow{y}=c$ (c scalar) and $\overrightarrow{a}\times \overrightarrow{y}=\overrightarrow{b}$ is
(a) ${{\left| \overrightarrow{a} \right|}^{2}}\left( c\overrightarrow{a}-\left( \overrightarrow{a}\times \overrightarrow{b} \right) \right)$
(b) ${{\left| \overrightarrow{a} \right|}^{2}}\left( c\overrightarrow{a}-\left( \overrightarrow{a}\times \overrightarrow{b} \right) \right)$
(c) $\dfrac{1}{{{\left| \overrightarrow{a} \right|}^{2}}}\left( c\overrightarrow{a}-\left( \overrightarrow{a}\times \overrightarrow{b} \right) \right)$
(d) $\dfrac{1}{{{\left| \overrightarrow{a} \right|}^{2}}}\left( c\overrightarrow{a}-\left( \overrightarrow{a}\times \overrightarrow{b} \right) \right)$

Hint: First do cross product by a to both the sides of equation after that apply vector law of three products which are \[\overrightarrow{a}\times \overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{b}\left( a.c \right)-\overrightarrow{c}\left( a.b \right)\]by using facts that \[\overrightarrow{a}.\overrightarrow{b}=\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|\cos \theta \] where $\theta $ is angle between two vectors and \[\overrightarrow{a}\times \overrightarrow{b}=\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|\sin \theta \] where $\theta $ is the angle between two vectors. Hence do further calculations to get the desired result.

Complete step-by-step answer:
In the dot product of two vectors $\overrightarrow{c},\overrightarrow{d}$ then we can say that,
$\overrightarrow{c}.\overrightarrow{d}=\left| \overrightarrow{c} \right|\left| \overrightarrow{d} \right|\cos \theta $
Here $\theta $ is the angle between two vectors and $\left| c \right|$ and $\left| d \right|$ are absolute values of vectors $\overrightarrow{c}$and \[\overrightarrow{d}\] . Now in the question we all know that $\overrightarrow{a},\overrightarrow{b}$ are perpendicular to each other then the angle between them is ${{90}^{\circ }}$, so in the equation,
$\overrightarrow{a}.\overrightarrow{b}=\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|\cos ({{90}^{o}})$
Now we know, $\cos {{90}^{\circ }}=0$, so $\overrightarrow{a}.\overrightarrow{b}=0$.
Now in the cross product of two vectors $\overrightarrow{c},\overrightarrow{d}$ then we can say that,
$\overrightarrow{c}\times \overrightarrow{d}=\left| c \right|\left| d \right|\sin \theta $
Here $\theta $ is the angle between two vectors, $\left| c \right|$ and $\left| d \right|$ are absolute values of vectors $\overrightarrow{c}$ and $\overrightarrow{d}$ . Now in the question we all know that $\overrightarrow{a},\overrightarrow{b}$ is perpendicular to each other then the angle between them is ${{90}^{\circ }}$, so we can write
$\overrightarrow{a}\times \overrightarrow{b}=\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|\sin ({{90}^{o}})$
We know, $\sin {{90}^{\circ }}=1$, so $\overrightarrow{a}\times \overrightarrow{b}=\left| \overrightarrow{a} \right|\left| \overrightarrow{b} \right|$.
Now from the question,
$\overrightarrow{a}.\overrightarrow{y}=c$ and $\overrightarrow{a}\times \overrightarrow{y}=\overrightarrow{b}$
For $\overrightarrow{a}\times \overrightarrow{y}=\overrightarrow{b}$ we cross multiply $\overrightarrow{a}$ on both sides we get,
$\overrightarrow{a}\times \left( \overrightarrow{a}\times \overrightarrow{y} \right)=\overrightarrow{a}\times \overrightarrow{b}$
Now here we will use formula, $\overrightarrow{a}\times \left( \overrightarrow{b}\times \overrightarrow{c} \right)=\overrightarrow{b}\left( \vec{a}.\vec{c} \right)-\overrightarrow{c}\left( \vec{a}.\vec{b} \right)$, so above equation can be written as
$\left( \overrightarrow{a}.\overrightarrow{y} \right)\overrightarrow{a}-\left( \overrightarrow{a}.\overrightarrow{a} \right)\overrightarrow{y}=\overrightarrow{a}\times \overrightarrow{b}$
We know $\overrightarrow{a}.\overrightarrow{a}={{\left| \overrightarrow{a} \right|}^{2}}$ and it is given that $\overrightarrow{a}.\overrightarrow{y}=c$, so above equation can be written as,
\[\begin{align}
  & c\overrightarrow{a}-{{\left| \overrightarrow{a} \right|}^{2}}\overrightarrow{y}=\overrightarrow{a}\times \overrightarrow{b} \\
 & \Rightarrow {{\left| \overrightarrow{a} \right|}^{2}}\overrightarrow{y}=c\overrightarrow{a}-\left( \overrightarrow{a}\times \overrightarrow{b} \right) \\
 & \Rightarrow \overrightarrow{y}=\dfrac{1}{{{\left| \overrightarrow{a} \right|}^{2}}}\left\{ c\overrightarrow{a}-\left( \overrightarrow{a}\times \overrightarrow{b} \right) \right\} \\
\end{align}\]
Hence the correct option is (c) or (d).

Note: Students should be careful while applying vector law of 3 products which is \[\overrightarrow{a}\times \overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{b}\left( a.c \right)-\overrightarrow{c}\left( a.b \right)\].
If they write the vector product as \[\overrightarrow{a}\times \overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{c}\left( a.b \right)-\overrightarrow{b}\left( a.c \right)\], they will get option (a) or (b) as the answer.
Students generally make mistake in solving the term, $\overrightarrow{a}.\overrightarrow{a}={{\left| \overrightarrow{a} \right|}^{2}}$.