The capacitance of a conductor is defined as the ratio of the charge given to the rise in the potential of the conductor.
$$
\begin{aligned}
& C=\frac{q}{V}=\frac{q^2}{W} \quad\left(V=\frac{W}{q}\right) \\
& C=\frac{\text { ampere }^2-\text { second }^2}{\mathrm{~kg}^2-\text { metre }^2 \text { second }^{-2}}
\end{aligned}
$$
Hence, dimensionally $\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^4 \mathrm{~A}^2\right]$
From Ohm's law, $V=I R$,
Hence, $R=\frac{V}{I}=\frac{\text { Volt }}{\text { Ampere }}=\frac{W}{q} \times \frac{t}{q}$
$$
\begin{aligned}
R & =\frac{F \times s \times t}{q^2}=\mathrm{kg} \frac{\mathrm{m}}{\mathrm{s}^2} \times \frac{\mathrm{m} \times \mathrm{s}}{\mathrm{A}^2 \times \mathrm{s}^2} \\
& =\mathrm{kg} \mathrm{m}^2 \mathrm{~s}^{-3} \mathrm{~A}^{-2}
\end{aligned}
$$
Dimensionally, $R=\left[\mathrm{ML}^2 \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]$
Hence, $C R=\left[\mathrm{M}^{-1} \mathrm{~L}^{-2} \mathrm{~T}^4 \mathrm{~A}^2\right]\left[\mathrm{ML}^2 \mathrm{~T}^{-3} \mathrm{~A}^{-2}\right]$
$$
=\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}\right]
$$