A metal of atomic mass 63.5 g mol^-1 crystallises in a face-centred cubic lattice with atomic radius 128 pm (atoms touch along the face diagonal).…

In an fcc lattice atoms touch along the face diagonal, so $\sqrt2\,a=4r$ and $a=4r/\sqrt2=2\sqrt2\,r=2\sqrt2\times128\ \mathrm{pm}=362.0\ \mathrm{pm}=3.620\times10^{-8}\ \mathrm{cm}$. The volume of the unit cell is $a^3=4.745\times10^{-23}\ \mathrm{cm^3}$. A perfect fcc cell contains $Z=4$ atoms, giving a density of $\frac{4\times63.5}{6.022\times10^{23}\times4.745\times10^{-23}}=8.89\ \mathrm{g\,cm^{-3}}$. With $0.5\%$ of the sites vacant, the effective number of atoms per cell is $4\times0.995=3.98$, so the density is reduced in proportion: $8.888\times0.995=8.84\ \mathrm{g\,cm^{-3}}$. Vacancies remove mass without changing the cell dimensions, so the density falls slightly below the ideal value; the lattice parameter itself is fixed by the atomic radius.