To solve this problem, we need to use the formula for the density of a crystal lattice.$\\$The density $\rho$ of a crystal is given by:$\\ \rho = \frac{Z \cdot M}{a^3 \cdot N_A} \\$Where:$\\$• $\rho = 8.92$ g cm$^{-3}$ is the density of copper.$\\$• $Z = 4$ is the number of atoms per unit cell for an fcc lattice.$\\$• $M$ is the molar mass of copper, which we need to find.$\\$• $a = 3.608 \times 10^{-8}$ cm is the edge length of the unit cell.$\\$• $N_A = 6.022 \times 10^{23}$ mol$^{-1}$ is Avogadro's number.$\\$Rearrange the formula to solve for $M: \\ M = \frac{\rho \cdot a^3 \cdot N_A}{Z} \\$Substitute the known values:$\\ M = \frac{8.92 \cdot (3.608 \times 10^{-8})^3 \cdot 6.022 \times 10^{23}}{4} \\$Calculate $a^3: \\ a^3 = (3.608 \times 10^{-8})^3 = 4.696 \times 10^{-23}$ cm$^3 \\$Substitute $a^3$ into the equation for $M: \\ M = \frac{8.92 \cdot 4.696 \times 10^{-23} \cdot 6.022 \times 10^{23}}{4} \\$Simplify the expression:$\\ M = \frac{8.92 \cdot 4.696 \cdot 6.022}{4}$ g mol$^{-1} \\$Calculate the numerator:$\\ 8.92 \cdot 4.696 \cdot 6.022 = 252.8 \\$Divide by 4:$\\ M = \frac{252.8}{4} = 63.2$ g mol$^{-1} \\$Therefore, the atomic mass of copper is approximately $63.1$ u, which corresponds to Option 1.