Step 1: Think of the diamond lattice as two interpenetrating face-centred-cubic lattices offset by one quarter of the body diagonal. The conventional cell holds $8$ atoms.
Step 2: The touching condition sets the radius. Nearest neighbours sit a distance $\tfrac{\sqrt3}{4}a$ apart, so the atomic radius is $r = \tfrac{\sqrt3}{8}a$, meaning the atoms are quite small relative to the roomy tetrahedral framework.
Step 3: Packing fraction is $\text{APF} = \dfrac{N\cdot \tfrac{4}{3}\pi r^3}{a^3}$ with $N = 8$. Writing $r$ in units of $a$ gives $\text{APF} = \dfrac{\pi\sqrt3}{16}$.
Step 4: Numerically $\dfrac{\pi\sqrt3}{16} \approx 0.3401$, i.e. about $34\%$. This is far below the $74\%$ of close-packed FCC/HCP because the tetrahedral bonding of diamond is open and directional rather than close-packed.\[\boxed{34\%}\]