Explanation

(c)
In fcc$\ \rightarrow$ contribution of each atom present at the corner $= \frac{1}{8}$
Contribution of each atom at the face centre $= \frac{1}{2}$
Hence, the total number of atoms in fcc
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \left( 8 \times \frac{1}{8} \right) + \left( 6 \times \frac{1}{2} \right) = 1 + 3 = 4$