Let $R_H=k h^a e^b \quad \dots(i)$

As, $R=\frac{V}{I}$

$\therefore {\left[R_H\right] } =\left[ ML ^2 T ^{-3} A ^{-2}\right]$

$\Rightarrow h =E \cdot t$

$\Rightarrow {[h] }=\left[ ML ^2 T ^{-1}\right]$

$e=I \cdot t \Rightarrow[e]=[ A \cdot T ]$

Substituting above values in Eq. $(i)$, we have

${\left[ ML ^2 T ^{-3} A ^{-2}\right] }=k\left[ ML ^2 T ^{-1}\right]^e[ AT ]^b$

$=k\left[ M ^a L ^{2 a} T ^{-a+b} A ^b\right]$

Equating dimensions, we get

$a=1 \text { and } b=-2$

Hence, $\quad R_H=k\left(\frac{h}{e^2}\right)$

So, dimensions of hall resistance are same as that of $\frac{h}{e^2}$.