What is the packing efficiency and number of atoms per unit cell...

In a face-centered cubic (FCC) unit cell, atoms are present at the $8$ corners and the centers of the $6$ faces.

Number of atoms from corners $= 8 \times \dfrac{1}{8} = 1$

Number of atoms from faces $= 6 \times \dfrac{1}{2} = 3$

Total number of atoms per unit cell ($Z$) $= 1 + 3 = 4$

Let $a$ be the edge length of the unit cell and $r$ be the radius of the atom. In an FCC lattice, atoms touch each other along the face diagonal.

Face diagonal $= \sqrt{2}a = 4r \Rightarrow a = 2\sqrt{2}r$

Volume of the unit cell $= a^3 = (2\sqrt{2}r)^3 = 16\sqrt{2}r^3$

Volume occupied by $4$ atoms $= 4 \times \dfrac{4}{3}\pi r^3 = \dfrac{16}{3}\pi r^3$

Packing efficiency $= \dfrac{\text{Volume occupied by atoms}}{\text{Volume of unit cell}} \times 100$

Packing efficiency $= \dfrac{\dfrac{16}{3}\pi r^3}{16\sqrt{2}r^3} \times 100 = \dfrac{\pi}{3\sqrt{2}} \times 100 \approx 74\ \%$

Answer: Packing efficiency $= 74\ \%$ and number of atoms per unit cell $= 4$