For the given close packed structure of a salt made of cation X and...

No. of $x$ $= 6 \times \dfrac{1}{2} = 3$ atom (at face centres)

No. of $y$ $= 8 \times \dfrac{1}{8} = 1$ atom (at all corners)

So, the formula of compound is ${x}_{3}y$

Now

If radius of $x$ is $r$ and radius of $y$ is $R$ then side length $a = 2R$

and face diagonal $\sqrt{2}a = 2R + 2r$

$r = \left(\sqrt{2} - 1\right)R$

Now Packing fraction $= \dfrac{volumeofatoms}{Volumeofunitcell}$

$= \dfrac{1 \times \dfrac{4}{3}{\pi R}^{3} + 3 \times \dfrac{4}{3}{\pi r}^{3}}{{a}^{3}}$

Substituting $a = 2R$ and $r = 0 . 414R$

$= \dfrac{\dfrac{4}{3}{\pi R}^{3}\left(1 + 3{\left(0 . 414\right)}^{3}\right)}{{\left(2R\right)}^{3}}$

$= \dfrac{4\pi}{3 \times 8}\left(1 + 0 . 07096 \times 3\right)$

$= \dfrac{\pi}{6} \times 1 . 212 = 0 . 63$