Atom $X$ occupies the fcc lattice sites as well as alternate tetrahedral voids of the same lattice. The packing efficiency (in$\%$) of the resultant solid is closest to

The correct option is C $35$
No. of atoms per unit cell :
$Z=8\left[\text{corner}\right]\frac{1}{8}+6\left[\text{Face}\right]\frac{1}{2}+4\left[\text{Alternate TV}\right]$

$Z=1+3+4$
$Z=8$

It is diamond like structure in which
$2r=\frac{\sqrt{3}a}{4}$

$\Rightarrow a=\left(\frac{8r}{\sqrt{3}}\right)$

Packing efficiency = (Z x Volume of single lattice point x 100) / volume of the unit cell
$Packingefficiency=\frac{8\times \frac{4}{3}\pi r^3}{{\left[\frac{8r}{\sqrt{3}}\right]}^3}\times 100=35\%\left(roughly\right)$