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:

$$\rho = \frac{Z \times M}{a^3 \times N_A}$$

$1 \, \text{pm} = 10^{-10} \, \text{cm}$

$$a = 400 \, \text{pm} = 400 \times 10^{-10} \, \text{cm} = 4 \times 10^{-8} \, \text{cm}$$

$$\rho = \frac{2 \times 100 \, \text{g/mol}}{(64 \times 10^{-24} \, \text{cm}^3) \times (6.022 \times 10^{23} \, \text{mol}^{-1})}$$ $$\rho = \frac{200 \, \text{g}}{64 \times 6.022 \times 10^{-24+23} \, \text{cm}^3}$$ $$\rho = \frac{200 \, \text{g}}{64 \times 6.022 \times 10^{-1} \, \text{cm}^3}$$ $$\rho = \frac{200 \, \text{g}}{385.408 \times 10^{-1} \, \text{cm}^3}$$ $$\rho = \frac{200 \, \text{g}}{38.5408 \, \text{cm}^3}$$ $$\rho \approx 5.188 \, \text{g/cm}^3$$

The calculated density is $5.188 \, \text{g/cm}^3$, which matches option (c).

The final answer is $\text{c) } 5.188 \, \text{g/cm}^3$.