Let  $\left[{\epsilon }_0\right]$  denote the dimensional formula of the permittivity of the vacuum and $\left[{\mu }_0\right]$ that of the permeability of the vacuum. If  $M=mass,$$L=length,$$T=time$$and$ $I=electric current$. 

Electrostatic Force, $F=\frac{1}{4\pi {ϵ}_0}\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}^2}$

$\text{ Permittivity of free space }{ϵ}_{\mathrm{o}}=\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{4\pi {\mathrm{Fr}}^2}$

$\text{ Thus dimensional formula of }\left[{ϵ}_0\right]=\frac{\left[\mathrm{AT}\right]\left[\mathrm{AT}\right]}{\left[{\mathrm{MLT}}^{-2}\right]\left[{\mathrm{L}}^2\right]}$

$\Rightarrow \left[{ϵ}_0\right]=\left[{\mathrm{M}}^{-1}{ \mathrm{L}}^{-3}{ \mathrm{T}}^4{ \mathrm{A}}^2\right]$

Magnetic Force, $F=\frac{{\mu }_0}{4\pi }\frac{m_1{m}_2}{{\mathrm{r}}^2}$

${\text{ Permeability of free space, μ}}_{\mathrm{o}}=\frac{4{\mathrm{\pi Fr}}^2}{m_1{m}_2}$

$\text{ Thus dimensional formula of }\left[{\text{μ}}_0\right]=\frac{\left[{\mathrm{MLT}}^{-2}\right]\left[{\mathrm{L}}^2\right]}{\left[\mathrm{AL}\right]\left[\mathrm{AL}\right]}$

$\Rightarrow \left[{ϵ}_0\right]=\left[{\mathrm{M}}^1{ \mathrm{L}}^1{ \mathrm{T}}^{-2}{ \mathrm{A}}^{-2}\right]$