Answer
$J = 3.27\times 10^5~A/m^2$
Work Step by Step
We can find the cross-sectional area of the cable:
$R = \frac{\rho~L}{A}$
$A = \frac{\rho~L}{R}$
$A = \frac{(2.75\times 10^{-8}~\Omega\cdot m)(1~km)}{0.150~\Omega}$
$A = 1.833\times 10^{-4}~m^2$
We can find the current density:
$J = \frac{i}{A}$
$J = \frac{60.0~A}{1.833\times 10^{-4}~m^2}$
$J = 3.27\times 10^5~A/m^2$
