Let $[\varepsilon_0]$ denote the dimensional formula of the permittivity of the vacuum, and $[\mu_0]$ that of the permeability of the vacuum. If $M = \text{mass}$, $L = \text{length}$, $T = \text{time}$ and $I = \text{electric current}$,

Step 1: Identify the given quantities and their dimensions.

The fundamental dimensions are mass $M$, length $L$, time $T$, and electric current $I$.

Step 2: Determine the dimensional formula of $[\varepsilon_0]$.

Use Coulomb's law for the electrostatic force between two charges.

$$F = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^2}$$

Rearrange the equation to express $\varepsilon_0$ in terms of the other variables.

$$\varepsilon_0 = \frac{q_1 q_2}{4\pi F r^2}$$

Express the dimensions of charge $q$, force $F$, and distance $r$.

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

Substitute these dimensions into the expression for $[\varepsilon_0]$.

$$[\varepsilon_0] = \frac{(I T)^2}{(M L T^{-2})(L)^2}$$ $$[\varepsilon_0] = M^{-1}L^{-3}T^{4}I^{2}$$

Step 3: Determine the dimensional formula of $[\mu_0]$.

Use the formula for the magnetic force per unit length between two parallel current-carrying conductors.

$$\frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d}$$

Rearrange the equation to express $\mu_0$.

$$\mu_0 = \frac{2\pi d F}{L I_1 I_2}$$

Since $d$ and $L$ both represent length, their dimensions cancel out.

$$[\mu_0] = \frac{[F]}{[I]^2}$$

Substitute the dimensions of force and current.

$$[\mu_0] = \frac{M L T^{-2}}{I^2}$$ $$[\mu_0] = M L T^{-2}I^{-2}$$

Step 4: Verify the relationship using the speed of light.

The speed of light in vacuum $c$ is related to $\varepsilon_0$ and $\mu_0$ by $c = \frac{1}{\sqrt{\mu_0 \varepsilon_0}}$.

$$[\varepsilon_0][\mu_0] = \left(M^{-1}L^{-3}T^{4}I^{2}\right)\left(M L T^{-2}I^{-2}\right)$$ $$[\varepsilon_0][\mu_0] = L^{-2}T^{2}$$

This matches the dimension of $[c^{-2}]$.

Answer: $[\varepsilon_0] = M^{-1}L^{-3}T^{4}I^{2}$ and $[\mu_0] = M L T^{-2}I^{-2}$