(a) Resistance of a uniform metallic conductor is directly proportional to its length $(l)$ and inversely proportional to the area of cross-section (A).i.e.,$R \propto 1$and$R \propto \frac{1}{A}$ Combining equations. (i) and (ii) we get or$R \propto \frac{l}{A}$$R = \rho \frac{l}{A}$where $\rho$ (rho) is a constant of proportionality and is called the electrical resistivity of the material of the conductor. Now, putting the units of $R, A$ and $l$ in the expression.$\rho = \frac{\Omega\,\text{m}^2}{\text{m}}$$= \Omega\text{-m}$Thus, the S.I unit of resistivity is Ohmmeter $(\Omega\text{-m})$. (b) GivenResistance $R = 0.04\ \Omega$Area of cross section $A = 1.4 \times 10^{-6}\ \text{m}^2$Length $l = 2\ \text{m}$Using the formula,$R = \rho \frac{l}{A}$Resistivity $\rho$ can be calculated as$P = \frac{R \times A}{l}$$\text{So resistivity} = \frac{0.04 \times 1.4 \times 10^{-6}\,\text{m}^2}{2\,\text{m}}$$= 2.8 \times 10^{-8}\,\Omega\text{-meter}$