A copper cable of cross section 3.0 mm² carries 15 A…

Step 1. The current density. $J = \dfrac{I}{A} = \dfrac{15\ \text{A}}{3.0 \times 10^{-6}\ \text{m}^2} = 5.0 \times 10^6\ \text{A/m}^2$.

Step 2. The field. $E = \rho J = 1.7 \times 10^{-8}\ \Omega\,\text{m} \times 5.0 \times 10^6\ \text{A/m}^2 = 0.085$ V/m.

Step 3. The voltage. $V = EL = 0.085\ \text{V/m} \times 40\ \text{m} = 3.4$ V.