In an FCC unit cell there are 4 atoms at the lattice sites and 8 tetrahedral voids. Occupying alternate tetrahedral voids contributes another 4 atoms. Total atoms per unit cell: $4+4=8$.

The constraint comes from the closer contact: along the body diagonal a tetrahedral void lies at one-quarter of the diagonal from a corner atom, so $\frac{a\sqrt{3}}{4}=2r$, giving $a=\frac{8r}{\sqrt{3}}$.

Packing efficiency $=\frac{8\cdot \frac{4}{3}\pi r^3}{a^3}\times 100=\frac{\frac{32\pi }{3}{r}^3}{{\left(\frac{8r}{\sqrt{3}}\right)}^3}\times 100$.

${\left(\frac{8r}{\sqrt{3}}\right)}^3=\frac{512r^3}{3\sqrt{3}}$. Therefore packing efficiency $=\frac{32\pi /3}{512/(3\sqrt{3})}\times 100=\frac{32\pi \sqrt{3}}{512}\times 100=\frac{\pi \sqrt{3}}{16}\times 100\approx 34\%$.

Closest to 35%. Option (B).