Answer
The diameter is $~0.92~mm$
Work Step by Step
We can write an expression for the electron current $i_e$ using the number of electrons that pass through a cross-section in a time $t$:
$N_e = i_e~t$
$i_e = \frac{N_e}{t}$
We can find the radius of the wire:
$i_e = n_e~A~v_d$
$\frac{N_e}{t} = n_e~\pi~r^2~v_d$
$r^2 = \frac{N_e}{n_e~\pi~v_d~t}$
$r = \sqrt{\frac{N_e}{n_e~\pi~v_d~t}}$
$r = \sqrt{\frac{1.0\times 10^{16}}{(5.80\times 10^{28}~m^{-3})~(\pi)~(8.0\times 10^{-4}~m/s)(320\times 10^{-6}~s)}}$
$r = 4.6\times 10^{-4}~m$
$r = 0.46~mm$
Since the diameter is twice the radius, the diameter is $0.92~mm$
