Correct answer is: (C)\(\cfrac{16}{3}\pi r^3\)
Volume occupied by one atom of radius ‘r’=\(\cfrac{4}{3}\pi r^3.\)
\(\therefore\)In fcc unit cell, there are 4 atoms present.
present in fcc unit cell = \(4\times\cfrac{4}{3}\pi r^3=\cfrac{16}{3}\pi r^3\)