= Solution
The reciprocal-space volume inside a spherical <Fermi surface> is $\mathcal V_{\mathrm F}=4\pi k_{\mathrm F}^3/3$. The first <Brillouin zone> has volume $(2\pi)^3$ divided by the primitive real-space cell volume:
$$
\mathcal V_{\mathrm{BZ}}=\frac{(2\pi)^3}{a^3/4}=\frac{32\pi^3}{a^3}.
$$
Since $k_{\mathrm F}^3=3\pi^2(4/a^3)$,
$$
\boxed{\mathcal V_{\mathrm F}=\frac{16\pi^3}{a^3}
=\frac12\mathcal V_{\mathrm{BZ}}.}
$$
Numerically,
$$
\begin{array}{c|rr}
&\mathrm{Cu}&\mathrm{Ag}\\\hline
\mathcal V_{\mathrm F}/\mathrm{m^{-3}}&1.05037\times10^{31}&7.28170\times10^{30}\\
\mathcal V_{\mathrm{BZ}}/\mathrm{m^{-3}}&2.10074\times10^{31}&1.45634\times10^{31}
\end{array}
$$
The half-zone volume reflects one electron per primitive cell and two spin states per Bloch wavevector: a completely filled single band would require two electrons per cell. Volume alone does not establish containment inside the polyhedral zone. Here the nearest reciprocal-lattice points are of type $(111)$, so their zone faces are a distance $\pi\sqrt3/a$ from the origin. The free-electron radius obeys
$$
\frac{k_{\mathrm F}}{\pi\sqrt3/a}=\frac{(12\pi^2)^{1/3}}{\pi\sqrt3}\simeq0.903<1,
$$
so these ideal monovalent spheres fit inside the first zone. Real band structure can distort a Fermi surface into zone-face contacts or necks; that is additional physics beyond the free sphere.
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