= Solution
A monovalent FCC metal supplies one conduction <electron> per atom and four atoms per conventional cell. Hence
$$
n_m=\frac4{(0.36\times10^{-9})^3}=8.57\times10^{28}\,\mathrm{m^{-3}}.
$$
With a one-carrier conductivity $\sigma=n_me\mu_m$,
$$
\boxed{\mu_m=\frac1{\rho_mn_me}=\frac1{(1.7\times10^{-8})(8.57\times10^{28})e}
=4.28\times10^{-3}\,\mathrm{m^2\,V^{-1}\,s^{-1}}.}
$$
This is about $84$ times smaller than the stated germanium <electron mobility>. Nevertheless the metal is vastly more conductive because its carrier density is about $3.7\times10^9$ times larger. Resistivity depends on both density and mobility; low resistivity alone does not establish high mobility or long scattering time.
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