Answer
$4.7\times10^{-4}m$
Work Step by Step
Use equation 18–3 to find the diameter. The cross-sectional area $A=\pi r^2=\frac{\pi d^2}{4}$.
$$R=\rho\frac{\mathcal{l}}{A}=\rho\frac{4 \mathcal{l}}{\pi d^2}$$
$$d=\sqrt{\frac{4 \mathcal{l}\rho}{\pi R}}$$
$$d=\sqrt{\frac{4(1.00m)(5.6\times10^{-8}\Omega \cdot m)}{\pi (0.32 \Omega)}}$$
$$=4.7\times10^{-4}m$$
