Answer
$\sigma = 7.50\times 10^6~(\Omega\cdot m)^{-1}$
Work Step by Step
We can find the current density:
$\sigma = \frac{J}{E}$
$J = \sigma~E$
$J = \frac{\sigma~V}{L}$
$J = \frac{(3.00\times 10^7~(\Omega\cdot m)^{-1})~(3.00~V)}{3.00\times 10^{-3}~m}$
$J = 3.00\times 10^{10}~A/m^2$
Since the three sections have the same current and the same cross-sectional area, the three sections have the same current density.
We can find the conductivity of section 2:
$\sigma = \frac{J}{E}$
$\sigma = \frac{J}{V/L}$
$\sigma = \frac{J~L}{V}$
$\sigma = \frac{(3.00\times 10^{10}~A/m^2)(1.00\times 10^{-3}~m)}{4.00~V}$
$\sigma = 7.50\times 10^6~(\Omega\cdot m)^{-1}$
