JEE Advanced 2022 · Paper 2, Chemistry question 15

Atoms per cell. fcc lattice sites give 4 atoms. Alternate tetrahedral voids give 4 of the 8, so the cell holds 8 atoms.

Atom radius. A tetrahedral void is a quarter of the way along the body diagonal, so an atom there touches the corner atom:
$$2r=\frac{\sqrt3a}4\implies r=\frac{\sqrt3a}8.$$

Packing efficiency.
$$\eta=\frac{8\cdot\frac43\pi r^3}{a^3}=\frac{32\pi}3\cdot\frac{3\sqrt3}{512}=\frac{\sqrt3\pi}{16}\approx0.34.$$
That is about 34%, closest to 35. (B).