Let $ {E_1}(r) $ , $ {E_2}(r) $ and $ {E_3}(r) $ be the respective electric fields at a distance $ r $ from a point charge $ Q $ , an infinitely long wire with constant linear charge density $ \lambda $ and an infinite plane with uniform surface charge density $ \sigma $ . If $ {E_1}({r_0}) = {E_2}({r_0}) = {E_3}({r_0}) $ at a distance $ {r_0} $ , then
(A) $ Q = 4\sigma \pi r_0^2 $
(B) $ {r_0} = \dfrac{\lambda }{{2\pi \sigma }} $
(C) $ {E_1}\left( {\dfrac{{{r_0}}}{2}} \right) = 2{E_2}\left( {\dfrac{{{r_0}}}{2}} \right) $
(D) $ {E_2}\left( {\dfrac{{{r_0}}}{2}} \right) = 4{E_3}\left( {\dfrac{{{r_0}}}{2}} \right) $

Hint Equate equal expressions in pairs. Find the relationship between the charge distributions at distance $ {r_0} $ and at distance $ \dfrac{{{r_0}}}{2} $ i.e. for point charge, find the relationship between $ {E_1}({r_0}) $ and $ {E_1}\left( {\dfrac{{{r_0}}}{2}} \right) $ . Likewise for the other charge distributions.

Formula used: $ {E_1}\left( {{r_0}} \right) = \dfrac{Q}{{4\pi {\varepsilon _o}r_0^2}} $ where $ Q $ is charge, $ {\varepsilon _o} $ is the permittivity of free space and $ {r_0} $ is the distance from the charge
$\Rightarrow {E_2}({r_0}) = \dfrac{\lambda }{{2\pi {\varepsilon _0}{r_0}}} $ and $ {E_3}({r_0}) = \dfrac{\sigma }{{2{\varepsilon _o}}} $ , where $ \lambda $ is the line charge density and $ \sigma $ is the surface charge density.

Complete step by step answer
Firstly, we write the equation of an electric field by a point charge $ Q $ at a distance $ {r_0} $ . This is given by
 $\Rightarrow {E_1}({r_0}) = \dfrac{Q}{{4\pi {\varepsilon _o}r_0^2}} $ where $ {\varepsilon _o} $ is the permittivity of free space.
Similarly, we write the equation of an electric field for an infinite line charge of density $ \lambda $ at a distance $ {r_0} $ . This is given by
 $\Rightarrow {E_2}({r_0}) = \dfrac{\lambda }{{2\pi {\varepsilon _o}{r_0}}} $
Finally,we write the equation of an electric field for an infinite surface charge of density $ \sigma $ at a distance $ {r_0} $ . This is written as
 $\Rightarrow {E_3}({r_0}) = \dfrac{\sigma }{{2{\varepsilon _o}}} $
According to the question, all three equations are equal. Thus, we could equate any pair so as to simplify and compare with the options.
For option A, we equate $ {E_1} $ and $ {E_2} $ . Doing this we get,
 $\Rightarrow \dfrac{Q}{{4\pi {\varepsilon _o}r_0^2}} = \dfrac{\sigma }{{2{\varepsilon _o}}} $
To make $ Q $ subject of the formula, multiply both sides by $ 4\pi {\varepsilon _o}r_0^2 $ , cancel out $ {\varepsilon _o} $ , and divide numerator and denominator by 2.
Hence, we get
 $\Rightarrow Q = 2\sigma \pi r_0^2 $ .
This is not the same as option A. Thus, option A is incorrect.
For option B, equating $ {E_1} $ and $ {E_3} $ we get
 $\Rightarrow \dfrac{\lambda }{{2\pi {\varepsilon _o}{r_0}}} = \dfrac{\sigma }{{2{\varepsilon _o}}} $
Cross multiplying and cancelling 2 and $ {\varepsilon _o} $ , we get
 $\Rightarrow \lambda = \sigma \pi {r_0} $
Rearranging the above equation we get,
 $\Rightarrow {r_0} = \dfrac{\lambda }{{\sigma \pi }} $ which is not equal to option B.
Therefore, option B is incorrect.
For option C, we need to find $ {E_1}\left( {\dfrac{{{r_0}}}{2}} \right) $ and $ {E_2}\left( {\dfrac{{{r_0}}}{2}} \right) $ .
We do this by simply replacing $ {r_0} $ with $ \dfrac{{{r_0}}}{2} $ in $ {E_1}({r_0}) = \dfrac{Q}{{4\pi {\varepsilon _o}r_0^2}} $ .
This gives
 $\Rightarrow {E_1}\left( {\dfrac{{{r_0}}}{2}} \right) = \dfrac{Q}{{4\pi {\varepsilon _o}\dfrac{{r_0^2}}{4}}} = 4\dfrac{Q}{{4\pi {\varepsilon _o}r_0^2}} \\
   \Rightarrow {E_1}\left( {\dfrac{{{r_0}}}{2}} \right) = 4{E_1}({r_0}) \\
  $
Similarly,
$\Rightarrow {E_2}\left( {\dfrac{{{r_0}}}{2}} \right) = \dfrac{\lambda }{{2\pi {\varepsilon _o}\dfrac{{{r_0}}}{2}}} = 2\dfrac{\lambda }{{2\pi {\varepsilon _o}{r_0}}} \\
   \Rightarrow {E_2}\left( {\dfrac{{{r_0}}}{2}} \right) = 2{E_2}({r_0}) \\
  $
This in turn, implies that
 $\Rightarrow \dfrac{1}{2}{E_2}\left( {\dfrac{{{r_0}}}{2}} \right) = {E_2}({r_0}) = {E_1}({r_0}) $ .
Substituting $ \dfrac{1}{2}{E_2}\left( {\dfrac{{{r_0}}}{2}} \right) $ for $ {E_1}({r_0}) $ in $ {E_1}\left( {\dfrac{{{r_0}}}{2}} \right) = 4{E_1}\left( {{r_0}} \right) $ we get,
 $ {E_1}\left( {\dfrac{{{r_0}}}{2}} \right) = 2{E_2}\left( {\dfrac{{{r_0}}}{2}} \right) $
This is equal to option (C).
Hence, the correct is option (C).

Note
Alternatively, in option C, from $ {E_2}\left( {\dfrac{{{r_0}}}{2}} \right) = 2{E_2}\left( {{r_0}} \right) $ we can say that
 $ {E_2}\left( {\dfrac{{{r_0}}}{2}} \right) = 2 \times \dfrac{1}{4} \times {E_1}\left( {\dfrac{{{r_0}}}{2}} \right) $ (since $ {E_2}({r_0}) = {E_1}({r_0}) $ ).
Simplifying further we get,
 $ {E_1}\left( {\dfrac{{{r_0}}}{2}} \right) = 2{E_2}\left( {\dfrac{{{r_0}}}{2}} \right) $ .