For the given close packed structure of a salt made of cation X and anion Y shown below (ions of only one…

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

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

So, the formula of compound is ${\mathrm{x}}_3\mathrm{y}$

Now

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

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

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

Now Packing fraction $=\frac{\mathrm{volume} \mathrm{of} \mathrm{atoms} }{\mathrm{Volume} \mathrm{of} \mathrm{unit} \mathrm{cell}}$

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

Substituting $\mathrm{a}=2 \mathrm{R}$ and $\mathrm{r}=0.414 \mathrm{R}$

$=\frac{\frac{4}{3}{\mathrm{\pi R}}^3\left(1+3{\left(0.414\right)}^3\right)}{{\left(2\mathrm{R}\right)}^3}$

$=\frac{4\mathrm{\pi }}{3\times 8} \left(1+0.07096\times 3\right)$

$=\frac{\mathrm{\pi }}{6}\times 1.212=0.63$