According to the NCERT text, the resistance $R$ of a conductor is given by $R = \frac{\rho l}{A}$, where $\rho$ is the resistivity, $l$ is the length, and $A$ is the cross-sectional area . Rearranging to solve for resistivity: $\rho = \frac{R A}{l}$. Substituting the given values ($R = 5.0 \ \Omega$, $A = 6.0 \times 10^{-7} \text{ m}^2$, $l = 15 \text{ m}$): $\rho = \frac{5.0 \times 6.0 \times 10^{-7}}{15} = \frac{30 \times 10^{-7}}{15} = 2.0 \times 10^{-7} \ \Omega\text{m}$. This matches the solution for Exercise 3.4 .