Let $[\epsilon_0]$ denote the dimensional formula of the permittivity of vacuum. If $M$ = mass, $L$ = length, $T$ = time and $A$ = electric current, then:

Step 1: State the relation using Coulomb's Law.

The electrostatic force $F$ between two charges $q_1$ and $q_2$ separated by a distance $r$ in vacuum is given by Coulomb's Law.

$$F = \frac{1}{4\pi \epsilon_0} \frac{q_1 q_2}{r^2}$$

Step 2: Express the permittivity of vacuum in terms of the other quantities.

Rearrange the formula to solve for $\epsilon_0$.

$$\epsilon_0 = \frac{q_1 q_2}{4\pi F r^2}$$

Since $4\pi$ is a dimensionless constant, the dimensional equation is written as follows.

$$[\epsilon_0] = \frac{[q^2]}{[F][r^2]}$$

Step 3: Determine the dimensions of charge, force, and distance.

Electric charge $q$ is the product of electric current and time.

$$[q] = [A T]$$ $$[q^2] = [A^2 T^2]$$

Force $F$ is the product of mass and acceleration.

$$[F] = [M L T^{-2}]$$

The square of the distance $r^2$ has the dimension of length squared.

$$[r^2] = [L^2]$$

Step 4: Substitute the dimensions to find $[\epsilon_0]$.

Substitute the individual dimensions back into the dimensional equation.

$$[\epsilon_0] = \frac{[A^2 T^2]}{[M L T^{-2}][L^2]}$$ $$[\epsilon_0] = [M^{-1} L^{-3} T^4 A^2]$$

Answer: The dimensional formula of the permittivity of vacuum is $[\epsilon_0] = [M^{-1} L^{-3} T^4 A^2]$.