Packing fraction in face-centered cubic unit cell is :

The correct option is B $0.7406$
Packing efficiency (P.E.) $=\frac{n\times \frac{4}{3}\pi r^3}{V}\times 100$

For a FCC lattice, $n=4$

Therefore, P.E. $=\frac{4\times \frac{4}{3}\times {(\frac{a}{2\sqrt{2}}}^3)}{a^3}=\frac{\pi }{3\sqrt{2}}\times 100=74\%$

Hence. in FCC, $74\%$ of the total volume is occupied by atoms.