Let [ε₀] denotes the dimensional formula of the permittivity of the vacuum and [μ₀] that of the permeability of the vacuum. If M = mass, L= length, T= time and I= electric current, then

The correct option is [μ₀] = MLT⁻²I⁻².

To find the dimensional formula of the permeability of vacuum (${\mu }_0$), we can use the formula for the magnetic force per unit length between two parallel, long current-carrying conductors separated by a distance $d$ in vacuum:
$\frac{F}{l}=\frac{{\mu }_0{I}_1{I}_2}{2\pi d}$

Rearranging the equation to solve for the permeability ${\mu }_0$:
${\mu }_0=\frac{2\pi d·F}{l·I_1{I}_2}$

Now, let's write the dimensional formulas for each term on the right-hand side:
- Constant $2\pi$ is dimensionless.
- Distance $\left[d\right]=L$
- Force $\left[F\right]=ML{T}^{-2}$
- Length $\left[l\right]=L$
- Currents $\left[I_1\right]=\left[I_2\right]=I$

Substituting these dimensions into the expression for ${\mu }_0$:
$\left[{\mu }_0\right]=\frac{L·\left(ML{T}^{-2}\right)}{L·I^2}$

Simplifying the dimensional equation:
$\left[{\mu }_0\right]=ML{T}^{-2}{I}^{-2}$

Thus, the dimensional formula of the permeability of vacuum is indeed $\left[{\mu }_0\right]=ML{T}^{-2}{I}^{-2}$.