Answer
$0.1045\;\rm V/m$
Work Step by Step
We know that the current density is given by
$$J=\sigma E=\dfrac{I}{A}$$
So,
$$I=\sigma E A$$
We are told that the current in both wires is the same, so
$$I_{\rm nichrome}=I_{\rm aluminum}$$
$$\sigma_{\rm nichrome} E_{\rm nichrome} A_{\rm nichrome}=\sigma_{\rm aluminum} E_{\rm aluminum} A_{\rm aluminum}$$
where $A=\pi r^2=\pi (D/2)^2=\pi D^2/4$
$$\sigma_{\rm nichrome} E_{\rm nichrome} \dfrac{\pi D^2_{\rm nichrome}}{4}=\sigma_{\rm aluminum} E_{\rm aluminum} \dfrac{\pi D^2_{\rm aluminum}}{4}$$
$$\sigma_{\rm nichrome} E_{\rm nichrome} D^2_{\rm nichrome} =\sigma_{\rm aluminum} E_{\rm aluminum} D^2_{\rm aluminum} $$
where $D_{\rm nichrome} =2D_{aluminum}$
$$\sigma_{\rm nichrome} E_{\rm nichrome} (2D_{\rm aluminum})^2 =\sigma_{\rm aluminum} E_{\rm aluminum} D^2_{\rm aluminum} $$
$$4\sigma_{\rm nichrome} E_{\rm nichrome} =\sigma_{\rm aluminum} E_{\rm aluminum} $$
Solving for $E_{nichrome}$;
$$ E_{\rm nichrome} =\dfrac{\sigma_{\rm aluminum} E_{\rm aluminum} }{4\sigma_{\rm nichrome}}$$
Plug the known (Table 30.2):
$$ E_{\rm nichrome} =\dfrac{(3.5\times 10^7)(0.0080)}{4(6.7\times 10^5)}$$
$$ E_{\rm nichrome} =\color{red}{\bf 0.1045}\;\rm V/m$$
