= Solution
Interpret “3 mm wire” as a circular wire of diameter $3\,\mathrm{mm}$, and take the current density as uniform. One conduction <Electron> per atom gives
$$
n=\frac{8940}{0.0635}N_A=8.48\times10^{28}\,\mathrm{m^{-3}},\qquad
A=\pi(1.5\times10^{-3})^2=7.07\times10^{-6}\,\mathrm{m^2}.
$$
Using $I=neA|v_d|$, the <drift velocity> corresponding to a current magnitude of $13\,\mathrm A$ is
$$
\boxed{|v_d|=\frac{13}{neA}=1.35\times10^{-4}\,\mathrm{m\,s^{-1}}}.
$$
It points opposite to conventional <electric current>. A simple distance scale over the $1/50=0.020\,\mathrm s$ cycle is $\boxed{|v_d|/f=2.71\,\mathrm{\mu m}}$, but this treats the magnitude as constant. For sinusoidal mains, <alternating-current drift excursion> gives zero net displacement after a full cycle.
If $13\,\mathrm A$ is the usual <root mean square> current, $v_{\rm pk}=\sqrt2|v_d|$ and the displacement amplitude and total absolute drift distance per cycle are
$$
\boxed{x_{\rm pk}=\frac{\sqrt2|v_d|}{2\pi f}=0.609\,\mathrm{\mu m}},\qquad
\boxed{s_{\rm cycle}=4x_{\rm pk}=2.44\,\mathrm{\mu m}}.
$$
If $13\,\mathrm A$ instead denotes peak current, these become $0.431\,\mathrm{\mu m}$ and $1.72\,\mathrm{\mu m}$. The paper does not state peak versus RMS; the characteristic drift is the boxed $1.35\times10^{-4}\,\mathrm{m\,s^{-1}}$ on whichever current convention is chosen. It also does not explicitly say diameter; taking $3\,\mathrm{mm}$ as the radius would reduce all these velocities and drift distances by four. None of these distances is the microscopic path travelled at roughly the <Fermi velocity> between scattering events; that path cannot be obtained from the supplied current alone.
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