Answer
4.2 mm.
Work Step by Step
Assume that we have exactly one meter of wire, carrying 35 A of current, dissipating 1.5 W of heat. Solve for the diameter.
The power dissipated is $I^2R$ and the resistance is $R=\frac{\rho\mathcal{l}}{A}$.
$$P=I^2\frac{\rho\mathcal{l}}{\pi(d/2)^2}$$
$$d=\sqrt{I^2\frac{4\rho\mathcal{l}}{P\pi}}=2I\sqrt{ \frac{\rho\mathcal{l}}{P\pi}}$$
$$d=2(35A)\sqrt{ \frac{(1.68\times10^{-8}\Omega\cdot m)(1.0m)}{(1.5W)\pi}}=4.2mm$$