Atom X occupies the fcc lattice sites as well as alternate tetrahedral voids of the same lattice. The…

Atom $\mathrm{X}$ occupies FCC lattice sites as well as alternate tetrahedral voids of FCC. In FCC, tetrahedral voids are $8$ (in a unit cell) Hence

Atom $\mathrm{X}$ in a unit cell (FCC lattice sites)

$=8\times \frac{1}{8}+6\times \frac{1}{2}=4$

Atom $\mathrm{X}$ in a unit cell (in T.V.) $=\frac{1}{2}\times 8=4$

Total atom$\mathrm{X}$  in one unit cell $=8$

For relation between a and r, since Tetrahedral void  forms at $\frac{1}{4}$th of body diagonal,

$\frac{\mathrm{a}\sqrt{3}}{4}=2\mathrm{r}$

$\mathrm{a}=\frac{8\mathrm{r}}{\sqrt{3}}$

Packing efficiency $=\frac{8\times \frac{4}{3}{\mathrm{\pi r}}^3}{{\mathrm{a}}^3}\times 100=\frac{8\times \frac{4}{3}{\mathrm{\pi r}}^3}{{\left(\frac{8\mathrm{r}}{\sqrt{3}}\right)}^3}\times 100=35\%$