Step 1: Approach
Use the packing efficiency of fcc, which is 74 percent, applied to the volume of the unit cell.
Step 2: Volume of the unit cell
\[ a=200\ \text{pm}=2\times10^{-8}\ \text{cm},\qquad a^3=8\times10^{-24}\ \text{cm}^3 \]
Step 3: Volume occupied
Packing fraction for fcc is $\dfrac{\pi}{3\sqrt2}=0.7405$.
\[ V_{occ}=0.7405\times8\times10^{-24}=5.92\times10^{-24}\ \text{cm}^3 \]
Step 4: Conclusion
Option (C), $8\times10^{-24}$ cm$^3$, is the whole cell, not only the atoms. The atoms alone occupy $5.92\times10^{-24}$ cm$^3$, so the answer is (B).
Final Answer:
The four atoms in the fcc cell fill $5.92\times10^{-24}\ \text{cm}^3$, option (B).
\[ \boxed{5.92\times10^{-24}\ \text{cm}^3} \]