= Solution
With one valence electron per atom, the electron <number density> is $n=4/a^3$. For a spin-degenerate, three-dimensional <free electron model> at zero temperature,
$$
k_{\mathrm F}=(3\pi^2n)^{1/3},\qquad E_{\mathrm F}=\frac{\hbar^2k_{\mathrm F}^2}{2m_e}.
$$
Substitution gives
$$
\begin{array}{c|rr}
&\mathrm{Cu}&\mathrm{Ag}\\\hline
n/\mathrm{m^{-3}}&8.4690\times10^{28}&5.8712\times10^{28}\\
k_{\mathrm F}/\mathrm{m^{-1}}&1.35858\times10^{10}&1.20240\times10^{10}\\
E_{\mathrm F}/\mathrm{eV}&7.032&5.508
\end{array}
$$
\b[Copper has the larger Fermi wavevector and Fermi energy because its smaller cell gives a larger electron density.] The energies are measured from the bottom of the free-electron dispersion, not from vacuum. Treating these real metals as monovalent free-electron systems is the model assumption, not a statement that their filled d bands are absent.
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