Atom X occupies the fcc lattice sites as well as alternate...

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

Atom $X$ in a unit cell (FCC lattice sites)

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

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

Total atom$X$ in one unit cell $= 8$

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

$\dfrac{a\sqrt{3}}{4} = 2r$

$a = \dfrac{8r}{\sqrt{3}}$

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