Answer
$5.1\times10^{-2}\Omega$.
Work Step by Step
Use equation 18–3 to find the resistance. The cross-sectional area $A=\pi r^2=\frac{\pi d^2}{4}$.
$$R=\rho\frac{\mathcal{l}}{A}=\rho\frac{4 \mathcal{l}}{\pi d^2}$$
$$R=(1.68\times10^{-8}\Omega \cdot m)\frac{4 (5.4m)}{\pi (1.5\times10^{-3}m)^2}$$
$$R=5.1\times10^{-2}\Omega$$
