Step 1: Identify the given values and the formula for density. The atomic mass ($M$) of the element is $100 \, \text{g/mol}$. The unit cell edge length ($a$) is $400 \, \text{pm}$. The crystal structure is Body-Centered Cubic (BCC), for which the number of atoms per unit cell ($Z$) is $2$. Avogadro's number ($N_A$) is $6.022 \times 10^{23} \, \text{mol}^{-1}$. The formula for the density ($\rho$) of a unit cell is: $$\rho = \frac{Z \times M}{a^3 \times N_A}$$ Step 2: Convert the unit cell edge length from picometers (pm) to centimeters (cm). $1 \, \text{pm} = 10^{-10} \, \text{cm}$ $$a = 400 \, \text{pm} = 400 \times 10^{-10} \, \text{cm} = 4 \times 10^{-8} \, \text{cm}$$ Step 3: Calculate the volume of the unit cell ($a^3$). $$a^3 = (4 \times 10^{-8} \, \text{cm})^3 = 4^3 \times (10^{-8})^3 \, \text{cm}^3 = 64 \times 10^{-24} \, \text{cm}^3$$ Step 4: Substitute the values into the density formula and calculate. $$\rho = \frac{2 \times 100 \, \text{g/mol}}{(64 \times 10^{-24} \, \text{cm}^3) \times (6.022 \times 10^{23} \, \text{mol}^{-1})}$$ $$\rho = \frac{200 \, \text{g}}{64 \times 6.022 \times 10^{-24+23} \, \text{cm}^3}$$ $$\rho = \frac{200 \, \text{g}}{64 \times 6.022 \times 10^{-1} \, \text{cm}^3}$$ $$\rho = \frac{200 \, \text{g}}{385.408 \times 10^{-1} \, \text{cm}^3}$$ $$\rho = \frac{200 \, \text{g}}{38.5408 \, \text{cm}^3}$$ $$\rho \approx 5.188 \, \text{g/cm}^3$$ Step 5: Compare the calculated density with the given options. The calculated density is $5.188 \, \text{g/cm}^3$, which matches option (c). The final answer is $\boxed{\text{c) } 5.188 \, \text{g/cm}^3}$.