The dimensional formula for the constant ${\varepsilon _0}$ is:
A) ${M^{ - 1}}{L^{ - 3}}{T^4}{A^2}$
B) $M{L^{ - 3}}{T^3}{A^2}$
C) $M{L^{ - 3}}{T^1}{A^2}$
D) ${M^2}{L^{ - 2}}{T^3}{A^{ - 2}}$

Hint: The electrostatic force between two charges separated by a distance is known to be proportional to the product of the two charges and inversely proportional to the square of the distance between the two charges. The corresponding constant of proportionality is termed as the permittivity of free space. The given constant ${\varepsilon _0}$ represents the permittivity of free space.

Formula used:
The electrostatic force between two charges is given by, $F = \dfrac{{{q_1}{q_2}}}{{4\pi {\varepsilon _0}{r^2}}}$ where ${q_1}$ and ${q_2}$ are the two charges, $r$ is the distance between these charges and ${\varepsilon _0}$ is the permittivity of free space.

Complete step by step answer:
Step 1: Express the relation for the electrostatic force between two charges to obtain a corresponding relation for the permittivity of free space.
The electrostatic force between two charges is expressed as $F = \dfrac{{{q_1}{q_2}}}{{4\pi {\varepsilon _0}{r^2}}}$ --------- (1)
where ${q_1}$ and ${q_2}$ are the two charges, $r$ is the distance between these charges and ${\varepsilon _0}$ is the permittivity of free space.
Expressing equation (1) in terms of the permittivity of free space we have
${\varepsilon _0} = \dfrac{{{q_1}{q_2}}}{{4\pi F{r^2}}}$ -------- (2)

Step 2: Express the dimensional formula for each quantity involved in equation (2).
The dimensional formula for force is expressed as $F \to \left[ {ML{T^{ - 2}}} \right]$ .
The dimensional formula for the two charges ${q_1}$ or ${q_2}$ can be expressed as $q \to \left[ {AT} \right]$ , ( $A$ is the dimension for current).
The dimensional formula for distance is expressed as $r \to \left[ L \right]$ .
Expressing equation (2) in terms of dimensions we get, ${\varepsilon _0} \to \dfrac{{\left[ {AT} \right]\left[ {AT} \right]}}{{\left[ {ML{T^{ - 2}}} \right]\left[ {{L^2}} \right]}} = \dfrac{{\left[ {{A^2}{T^2}} \right]}}{{\left[ {M{L^3}{T^{ - 2}}} \right]}}$
On further simplifying, we obtain the dimensional formula for the permittivity of free space to be${\varepsilon _0} \to \left[ {{M^{ - 1}}{L^{ - 3}}{A^2}{T^4}} \right]$.

So the correct option is A.

Note: The dimensional formula for distance is length $\left[ L \right]$ , but for the square of the distance between the two charges, we substitute $\left[ {{L^2}} \right]$ when expressing equation (2) in terms of the dimensions of the quantities involved in it. The constant $4\pi $ in equation (2) does not have a dimensional formula. The charge can be taken as the product of current and time, so we have its dimensional formula as $q \to \left[ {AT} \right]$ .