Step 1: Work with a unit cube.
Let the lattice parameter be \( a = 1 \). In FCC, atoms touch along the face diagonal, so \( a\sqrt{2} = 4R \), giving \( R = \frac{\sqrt{2}}{4} \approx 0.3536 \).
Step 2: Find how much atom volume sits inside that cube.
An FCC cell has 8 corner atoms counted one-eighth each and 6 face atoms counted one-half each, giving \( 1 + 3 = 4 \) whole atoms. One atom has volume \( \frac{4}{3}\pi R^3 = \frac{4}{3}\pi(0.3536)^3 \approx 0.1851 \), so 4 atoms occupy about \( 0.7405 \).
Step 3: Compare to the cell volume.
Since the cube volume is \( 1^3 = 1 \):
\[ \text{APF} = \frac{0.7405}{1} \approx 0.74 \]
Step 4: Final Answer.
So close to 74 percent of the FCC cell is filled with atoms, the densest possible packing for equal spheres. \[ \boxed{\text{APF} \approx 0.74} \]
Therefore, option (C) is correct.