Step 1: Work with a unit cube.
Take the lattice parameter as \( a = 1 \). In BCC, atoms touch along the body diagonal, so \( a\sqrt{3} = 4R \), giving \( R = \frac{\sqrt{3}}{4} \approx 0.4330 \).
Step 2: Find the atom volume actually packed into that cube.
A BCC cell holds 8 corner atoms counted one-eighth each plus 1 full body-centre atom, so 2 whole atoms sit inside the cell. One atom has volume \( \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (0.4330)^3 \approx 0.3401 \), so the 2 atoms together occupy about \( 0.6802 \).
Step 3: Compare filled volume to cell volume.
Since the cube volume is \( 1^3 = 1 \):
\[ \text{APF} = \frac{0.6802}{1} \approx 0.68 \]
Step 4: Final Answer.
So about 68 percent of the BCC cell is filled with atoms, leaving 32 percent empty. \[ \boxed{\text{APF} \approx 0.68} \]
Therefore, option (B) is correct.