The unit of permittivity of free space, ${\mathrm{\epsilon }}_{\mathrm{o}}\mathrm{is}$

By Coulomb's law, the electrostatic force

$\mathrm{F}=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}\times \frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}^2}\Rightarrow {\mathrm{\epsilon }}_{\mathrm{o}}=\frac{1}{4\mathrm{\pi }}\times \frac{{\mathrm{q}}_1{\mathrm{q}}_2}{{\mathrm{r}}^2\mathrm{F}}$

Substituting the units of q, r and F, we obtain unit of

${\mathrm{\epsilon }}_{\mathrm{o}}\mathrm{\hspace{0.17em}}=\mathrm{\hspace{0.17em}}\frac{\mathrm{coulomb}\times \mathrm{coulomb}}{\mathrm{newton}-{\left(\mathrm{metre}\right)}^2}\mathrm{\hspace{0.17em}}=\mathrm{\hspace{0.17em}}\frac{{\left(\mathrm{coulomb}\right)}^2}{\mathrm{newton}-{\left(\mathrm{metre}\right)}^2}$

$={\mathrm{C}}^2/\mathrm{N}-{\mathrm{m}}^2$

Method 2: $\mathrm{F}=\frac{1}{4{\mathrm{\pi \epsilon }}_{\mathrm{o}}}.\frac{{\mathrm{Q}}_1{\mathrm{Q}}_2}{{\mathrm{r}}^2}\Rightarrow {\mathrm{\epsilon }}_{\mathrm{o}}\propto \frac{{\mathrm{Q}}^2}{\mathrm{F}\times {\mathrm{r}}^2}$

So ${\mathrm{\epsilon }}_{\mathrm{o}}$ has units of ${\mathrm{coulomb}}^2/\mathrm{newton}-{\mathrm{m}}^2$