(a) Resistance of a uniform metallic conductor is directly proportional to its length $\left(\mathit{l}\right)$ and inversely proportional to the area of cross-section (A).
i.e.,
$\mathrm{R}\propto 1$
and
$\mathrm{R}\propto \frac{1}{\mathrm{A}}$
Combining equations. (i) and (ii) we get or
$\mathrm{R}\propto \frac{\mathit{l}}{\mathrm{A}}$
$\mathrm{R}=\mathit{\rho }\frac{\mathit{l}}{\mathrm{A}}$
where $\mathit{\rho }$ (rho) is a constant of proportionality and is called the electrical resistivity of the material of the conductor. Now, putting the units of $\mathit{R},\mathit{A}$ and $\mathit{l}$ in the expression.
$\mathit{\rho }=\frac{\mathit{\Omega }{\mathit{m}}^2}{\mathit{m}}$
$=\mathit{\Omega }-\mathit{m}$
Thus, the S.I unit of resistivity is Ohmmeter $\left(\mathit{\Omega }-\mathit{m}\right)$.
(b) Given
Resistance $\mathrm{R}=0.04\mathrm{ohm}$
Area of cross section $\mathrm{A}=1.4\times {10}^{-6}{\mathrm{m}}^2$
Length $\mathit{l}=2\mathrm{m}$
Using the formula,
$\mathrm{R}=\mathit{\rho }\frac{\mathit{l}}{\mathrm{A}}$
Resistivity $\mathit{\rho }$ can be calculated as
$\mathrm{P}=\left(\mathrm{R}\times \mathrm{A}\right)∕\mathrm{l}$