To find: The combination of $\varepsilon_{0}$ and $\mu_{o}$ which has dimensional formula of electrical resistance.

Method - 1

Dimensional formula of resistance:

$\lbrack R\rbrack = \left\lbrack M^{- 1}{\text{\ }L}^{2}{\text{\ }T}^{- 3}{\text{\ }A}^{2} \right\rbrack$

Dimensions formula of $\varepsilon_{0}$:

$\varepsilon_{0} = \left\lbrack M^{- 1}{\text{\ }L}^{- 3}{\text{\ }T}^{4}{\text{\ }A}^{2} \right\rbrack$

(by Coulomb's law)

Dimensional formula of $\mu_{0}$:

$\mu_{0} = \left\lbrack M^{1}{\text{\ }L}^{1}{\text{\ }T}^{- 2}{\text{\ }A}^{- 2} \right\rbrack$

Let the required combination be:

$R = \left\lbrack \varepsilon_{0} \right\rbrack^{x}\left\lbrack \mu_{0} \right\}^{y}$

Putting dimensions or, $R,\varepsilon_{0}\ \&{\ \mu}_{0}$ from equations

$\left\lbrack M^{1}L^{2}T^{- 3}A^{- 2} \right\rbrack = \left\lbrack M^{- 1}L^{- 3}T^{4}A^{2} \right\rbrack^{x}\left\lbrack M^{1}L^{1}T^{- 2}A^{- 2} \right\rbrack^{y}$

$\left\lbrack M^{1}L^{2}T^{- 3}A^{- 2} \right\rbrack = \lbrack M\rbrack^{- x + y}\lbrack L\rbrack^{- 3x + y}\lbrack T\rbrack^{4x - 2y}\lbrack A\rbrack^{2x - 2y}$

Comparing the exponents:

$\begin{matrix} - x + y & \ = 1 \\ - 3x + y & \ = 2 \\ 4x - 2y & \ = - 3 \\ 2x - 2y & \ = - 2 \\ \end{matrix}$

Solving above set of linear equations, we get:

$x = - \frac{1}{2},$

$y = \frac{1}{2}$

Putting these values in equation (iv), we get the required expression:

$R = \sqrt{\frac{\mu_{0}}{\varepsilon_{0}}}$

Method - 2

$\sqrt{\frac{\mu_{0}}{\varepsilon_{0}}} = \frac{\sqrt{\mu_{0}\ }\sqrt{\varepsilon_{0}}}{\varepsilon_{0}} = \frac{1}{\varepsilon_{0}c}\ \lbrack c = velocity\ of\ light\rbrack$

$= \frac{Fr^{2}}{q^{2}c} = \frac{work \times \left( \frac{r}{c} \right)}{q^{2}} = \frac{Vqt}{q^{2}}\$

$\ \left\lbrack \because I = \frac{q}{t}\ and\ F = \frac{1}{4\pi\varepsilon_{0}}\frac{q^{2}}{r^{2}} \right\rbrack$

$= \frac{V \times t}{I \times t} = \frac{V}{I} = R$