Calculate the number of atoms in 5.4 g metal forming fcc structure $\left[\varrho \times a^3=7 \cdot 2 \times 10^{-22} \mathrm{~g}\right]$

Given $\varrho a^3 = 7.2 \times 10^{-22}$ g and FCC structure with $Z = 4$ atoms per unit cell, rearrange the density formula $\varrho = \frac{ZM}{N_A a^3}$ to solve for molar mass: $\varrho a^3 = \frac{ZM}{N_A}$.

Substitute values using $N_A = 6 \times 10^{23}$ mol$^{-1}$: $7.2 \times 10^{-22} = \frac{4M}{6 \times 10^{23}}$.

Solving yields $M = \frac{7.2 \times 10^{-22} \times 6 \times 10^{23}}{4} = 108$ g/mol.

The number of moles in 5.4 g is $n = \frac{5.4}{108} = 0.05$ mol.

Thus the number of atoms is $0.05 \times 6 \times 10^{23} = 3.0 \times 10^{22}$, corresponding to option A.