Answer
$\rho = 5.0\times 10^{-8}~\Omega~m$
Work Step by Step
We can find the resistivity of the wire:
$\rho = \frac{E}{J}$
$\rho = \frac{E}{I/A}$
$\rho = \frac{E~A}{I}$
$\rho = \frac{E~\pi~r^2}{I}$
$\rho = \frac{(0.085~V/m)~(\pi)~(1.5\times 10^{-3}~m)^2}{12~A}$
$\rho = 5.0\times 10^{-8}~\Omega~m$
