= Solution
Treat <sodium> as monovalent, with one mobile <Electron> per atom. Its <mass density> is $\rho_m=0.97\,\mathrm{g\,cm^{-3}}=970\,\mathrm{kg\,m^{-3}}$, and its molar <mass> is $M=0.023\,\mathrm{kg\,mol^{-1}}$. Using the <Avogadro constant>,
$$
n=\frac{\rho_mN_A}{M}=\frac{970}{0.023}(6.022\times10^{23})\,\mathrm{m^{-3}}
\simeq2.54\times10^{28}\,\mathrm{m^{-3}}.
$$
The <electron mobility> $\mu=e\tau/m_e$ relates the magnitude of <drift velocity> to <electric field>. The <Drude model> then gives
$$
\boxed{\sigma=ne\mu=(2.54\times10^{28})(1.602\times10^{-19})(5.2\times10^{-3})
\simeq2.1\times10^7\,\mathrm{S\,m^{-1}}.}
$$
Using $m_e=9.109\times10^{-31}\,\mathrm{kg}$,
$$
\boxed{\tau=\frac{m_e\mu}{e}\simeq2.96\times10^{-14}\,\mathrm s\simeq3.0\times10^{-14}\,\mathrm s.}
$$
The inferred time is the <momentum relaxation time>. Identifying it with the mean time between collisions is the intended simple isotropic <Drude model> assumption; small-angle scattering need not relax directed <momentum> in a single event.
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