Solution:
Key ldea : Substitute the units for all the quantities involved in an expression written for permittivity of free space.
By Coulomb's law, the electrostatic force
$$
\begin{aligned}
&F=\frac{1}{4 \pi \varepsilon_{0}} \times \frac{q_{1} q_{2}}{r^{2}} \\
&\Rightarrow \varepsilon_{0}=\frac{1}{4 \pi} \times \frac{q_{1} q_{2}}{r^{2} F}
\end{aligned}
$$
Substituting the units for \(q, r\) and \(F\), we obtain unit of
$$
\begin{aligned}
&\varepsilon_{0}=\frac{\text { coulomb } \times \text { coulomb }}{\text { newton }-(\text { metre })^{2}} \\
&=\frac{(\text { coulomb })^{2}}{\text { newton }-(\text { metre })^{2}} \\
&=C^{2} / N-m^{2}
\end{aligned}
$$
