A given element A has a face centred unit cell in which A is absent from two of the corners.
The packing fraction is :

The correct options are
B $\frac{5\pi }{16\sqrt{2}}$
F $\frac{5\pi }{16\sqrt{2}}$
$\text{Contribution for a corner particle is}=\frac{1}{8}$
$\text{Contribution for a face centre particle is}=\frac{1}{2}$

Total no. of particles :

$Z_{eff}=(6\times \frac{1}{8})+(6\times \frac{1}{2})=\frac{15}{4}$
For a FCC unit cell :
$\text{Edge length}\left(a\right)=2\sqrt{2}\times r$

Here r is the radius of particle A
$\text{Total volume}V=a^3=(2\sqrt{2}\times r{)}^{3}V=16\sqrt{2}\times r^3$
$\text{Volume occupied}\left(V_o\right)=Z_{eff}\times \frac{4}{3}\times \pi \times r^3{V}_o=\frac{15}{4}\times \frac{4}{3}\times \pi \times r^3$
$\text{Packing Fraction (P.F)}=\frac{V_o}{V}P.F=\frac{\frac{15}{4}\times \frac{4}{3}\times \pi \times r^3}{16\sqrt{2}\times r^3}$

$P.F=\frac{5\pi }{16\sqrt{2}}$