#### Answer

$0.104$ Volts/meter

#### Work Step by Step

Recall that $J = \sigma\vec{E}, \quad\rightarrow\quad I = A\sigma\vec{E}$
If the current is the same in both wires, $I_{nichrome} = I_{aluminum}$, then we have the equation:
$ A_n\sigma_n\vec{E}_n = A_A\sigma_A\vec{E}_A$
Table 27.2 gives the conductivity for nichrome as $\sigma_n = 6.7\times10^5\Omega$m and for aluminum as $\sigma_A = 3.5\times10^7\Omega$ m
Solving for $\vec{E}_n$...
$\displaystyle \vec{E}_n = \frac{A_A\sigma_A\vec{E}_A}{A_n\sigma_n} = \frac{(\pi(0.0005)^2)(3.5\times10^7)(0.008)}{(\pi(0.001)^2)(6.7\times10^5)} = 0.104$ Volts/meter