Answer
$7.23\;\rm mm$
Work Step by Step
The current is the same in both wires according to the current conservation principle.
Hence,
$$I_1=I_2=I\tag 1$$
However, the current density is not the same in both cases.
$$J_1=\dfrac{I_1}{A_1}=\sigma_1E_1\tag 2$$
and hence,
$$J_2=\dfrac{I_2}{A_2}=\sigma_2E_2\tag 3$$
where $_1$ is for the left wire (Aluminum) and $_2$ is for the right wire (Nichrome).
Divide (3) by (2),
$$\dfrac{\dfrac{I_2}{A_2}}{\dfrac{I_1}{A_1}}=\dfrac{\sigma_2E_2}{\sigma_1E_1}$$
where $I_1=I_2$,
$$\dfrac{A_1}{A_2}=\dfrac{\sigma_2E_2}{\sigma_1E_1}$$
where $A=\pi r^2=\pi (D/2)^2=\pi D^2/4$ where $D$ is the diameter of the wire,
$$\dfrac{\pi D_1^2/4}{\pi D_2^2/4}=\dfrac{\sigma_2E_2}{\sigma_1E_1}$$
$$\dfrac{ D_1^2 }{ D_2^2 }=\dfrac{\sigma_2E_2}{\sigma_1E_1}$$
Hence,
$$D_2^2=\dfrac{\sigma_1E_1}{\sigma_2E_2}D_1^2$$
And since we need the electric field strength to be the same in both wires, $E_1=E_2$;
$$D_2 =\sqrt{\dfrac{\sigma_1 }{\sigma_2 }}D_1$$
Plug the known;
$$D_2 =\sqrt{\dfrac{(3.5\times 10^7)}{(6.7\times 10^5)}}(1)$$
$$D_2=\color{red}{\bf 7.23}\;\rm mm$$
