= Solution
The nearest-neighbor displacement is $a[111]/4$, of length $a\sqrt3/4$. Touching <hard spheres> therefore obey
$$
2r_{\rm atom}=\frac{\sqrt3a}{4},\qquad
r_{\rm atom}=\frac{\sqrt3a}{8}.
$$
There are eight atoms per conventional cubic cell, giving
$$
\boxed{f_{\rm diamond}=
\frac{8(4\pi r_{\rm atom}^3/3)}{a^3}
=\frac{\pi\sqrt3}{16}\simeq0.3401.}
$$
For monatomic FCC close packing, four atoms per cell touch along a face diagonal, so $r_{\rm atom}=a/(2\sqrt2)$ and
$$
\boxed{f_{\rm ccp}=\frac{\pi}{3\sqrt2}\simeq0.7405.}
$$
Thus the <diamond> <packing fraction> is much smaller. The tetrahedral network leaves substantial open volume. Although its <Bravais lattice> is FCC, the two-atom motif and the shorter hard-sphere contact distance prevent it from inheriting the FCC close-packing fraction.
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