= Solution
In a conventional <face-centered cubic lattice> cell of side $a_F$, touching nearest-neighbour <hard spheres> lie along a face diagonal. Four radii span that diagonal, so $\sqrt2a_F=4r$ and $a_F=2\sqrt2r$. In a conventional <body-centered cubic lattice> cell, the contact lies on the body diagonal: $\sqrt3a_B=4r$, giving $a_B=4r/\sqrt3$. Therefore
$$
\boxed{\frac{V_F}{V_B}=\left(\frac{2\sqrt2r}{4r/\sqrt3}\right)^3=\frac{3\sqrt6}{4}=1.837}.
$$
These are the nonprimitive cubic <unit cells> requested. They contain four and two atoms respectively, so their volume ratio should not be confused with the volume-per-atom ratio, which is half as large, $3\sqrt6/8\simeq0.9186$. The common sphere radius cancels.
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