Answer
$\sigma = 2.0\times 10^6~(\Omega\cdot m)^{-1}$
Work Step by Step
We can find the conductivity:
$\sigma = \frac{J}{E}$
$\sigma = \frac{i/A}{V/L}$
$\sigma = \frac{i~L}{V~A}$
$\sigma = \frac{(4.0~A)(1.0~m)}{(2.0~V)(1.0\times 10^{-6}~m^3)}$
$\sigma = 2.0\times 10^6~(\Omega\cdot m)^{-1}$
