Solution:

The answer is (B). We observe that the solid of revolution obtained by rotating a rhombus about one of its sides can be seen as a cone superimposed on a "hollowed-out" cylinder. The region of space that is hollowed out of the cylinder is equal to the cone above it, consequently the required volume $V$ is equivalent to the volume of a cylinder having as height the side of the rhombus $(8~\mathrm{cm})$ and as radius the height of the rhombus (half the side, that is $4~\mathrm{cm}$). The calculation then gives $V=\pi \cdot 4^{2} \cdot 8=128 \pi$