A metal forms fcc unit cell with unit edge length 400\ pm ....

Edge length of the unit cell, $a = 400\ \text{pm} = 4 \times 10^{-8}\ \text{cm}$

Volume of the unit cell, $V = a^3 = (4 \times 10^{-8}\ \text{cm})^3 = 64 \times 10^{-24}\ \text{cm}^3$

Mass of one unit cell = Volume $\times$ Density = $64 \times 10^{-24}\ \text{cm}^3 \times 12.0\ \text{g/cm}^3 = 768 \times 10^{-24}\ \text{g}$

Number of unit cells in $0.6\ \text{g}$ of metal = $\dfrac{0.6}{768 \times 10^{-24}} = \dfrac{600}{768} \times 10^{21} = 0.78125 \times 10^{21}$

Since the metal forms an fcc unit cell, the number of atoms per unit cell is $4$.

Total number of atoms = $4 \times 0.78125 \times 10^{21} = 3.125 \times 10^{21}$

Answer: $3.125 \times 10^{21}$