Right answer is (c) 3.011 x 10^23
For explanation: Atomic mass of Al = 27g/mole (contains 6.022 x 10^23 Al atoms)
Since it exhibits FCC, there are 4 Al atoms/unit cell.
If 27g Al contains 6.022 x 10^23 Al atoms then 54g Al contains 1.2044 x 10^24atoms.
Thus, if 1 unit cell contains 4 Al atoms then number of unit cells containing 1.2044 x 10^24 atoms=(1.2044 x 10^24 x 1)/4 = 3.011× 10^23 unit cells.