To solve this problem, we need to find the edge length of a unit cell of Lithium, which has a bcc (body-centered cubic) structure.$\\$Given:$\\$• Density of Lithium $(\rho) = 520$ kg m$^{-3} = 0.520$ g cm$^{-3} \\$• Atomic mass of Lithium $(M) = 6.94$ g mol$^{-1} \\$• Avogadro's number $(N_A) = 6.02 \times 10^{23}$ mol$^{-1} \\$In a bcc structure, there are 2 atoms per unit cell.$\\$The formula for density is:$\\ \rho = \frac{Z \cdot M}{a^3 \cdot N_A} \\$Where:$\\$• $Z = 2$ (number of atoms per unit cell for bcc)$\\$• $a =$edge length of the unit cell in cm$\\$Rearranging the formula to solve for $a: \\ a^3 = \frac{Z \cdot M}{\rho \cdot N_A} \\$Substitute the known values:$\\ a^3 = \frac{2 \cdot 6.94}{0.520 \cdot 6.02 \times 10^{23}} \\ a^3 = \frac{13.88}{3.1304 \times 10^{23}} \\ a^3 = 4.435 \times 10^{-23}$ cm$^3 \\$Taking the cube root to find $a: \\ a = (4.435 \times 10^{-23})^{1/3} \\ a \approx 3.52 \times 10^{-8}$ cm$\\$Convert $a$ from cm to pm:$\\ 1$ cm$= 10^{10}$ pm$\\ a \approx 3.52 \times 10^{-8} \times 10^{10}$ pm$\\ a \approx 352$ pm$\\$Therefore, the edge length of the unit cell of Lithium is $352$ pm, \\ which corresponds to Option 2.