= Solution
For hard spheres of radius r in a <face-centered cubic lattice>, contact occurs along a face diagonal: $4r=\sqrt2a$. Four spheres belong to a conventional cell, so the <hard-sphere cubic packing fractions> give
$$
\boxed{\eta_{\rm FCC}=\frac{4(4\pi r^3/3)}{a^3}=\frac{\pi}{3\sqrt2}=0.7405.}
$$
For a <body-centered cubic lattice>, contact occurs along a body diagonal, $4r=\sqrt3a$, and there are two spheres per conventional cell:
$$
\boxed{\eta_{\rm BCC}=\frac{2(4\pi r^3/3)}{a^3}=\frac{\sqrt3\pi}{8}=0.6802.}
$$
Their ratio $8/(3\sqrt6)=1.0887>1$ proves that FCC packs more densely. Coordination alone would not establish the occupied volume fraction.
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