= Solution
For a spherical <Fermi surface> inside a <primitive cubic lattice> zone, the shortest reciprocal vector has magnitude $G=2\pi/a$. The nearest translated Fermi sphere is separated by the <electron-phonon Umklapp threshold>
$$
q_U=G-2k_F=\frac{0.6\pi}{a}
\simeq3.77\times10^9\,\mathrm{m^{-1}}.
$$
A phonon of this wavevector can connect nearly opposite-velocity Fermi states using $\mathbf k+\mathbf q=\mathbf k'+\mathbf G$, transferring reciprocal-lattice momentum to the crystal and efficiently relaxing electrical current. With the supplied sound speed, its thermal scale is
$$
\boxed{T_U\sim\frac{\hbar v_s q_U}{k_B}
\simeq86\,\mathrm K,\quad\text{of order }10^2\,\mathrm K.}
$$
For $T\ll T_U$ such <phonons> have exponentially small occupation. As they become populated, <Umklapp scattering> can make the <electrical conductivity> fall rapidly. Lower-energy normal scattering and residual disorder still give resistance below this scale; this is a crossover estimate, not a sharp transition. The numerical estimate extrapolates the linear acoustic dispersion to $q_U$ and assumes the spherical <Fermi surface> implicit in giving one $k_F$.
Using $2k_F$ alone as the phonon wavevector misses this mechanism: direct backscattering would need $1.4\pi/a$, beyond the first-zone edge along this direction, while the reciprocal-lattice transfer permits it with $0.6\pi/a$.
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