In a face-centred cubic lattice of an element, atoms touch along the face diagonal. If the edge length of the unit cell is $a$, the atomic radius $r$ is related as:

In an FCC lattice, atoms at the corners touch atoms at the face centres along the face diagonal. The face diagonal of a cube of edge $a$ has length $a\sqrt{2}$. Along that diagonal sit (corner atom) + (full face-centre atom) + (corner atom). Total length covered by atomic radii: $r + 2r + r = 4r$. So: $4r = a\sqrt{2} \Rightarrow r = \dfrac{a\sqrt{2}}{4}$. Compare with **BCC** (atoms touch along the body diagonal): $4r = a\sqrt{3}$, giving $r = \dfrac{a\sqrt{3}}{4}$ — option C, which is the BCC trap. **Simple cubic**: $r = a/2$ (option A), atoms touch along the edge.