Percentage of free space in a body centered cubic unit cell is -

In a body centred cubic (bcc) unit cell the atoms touch along the body diagonal: $\sqrt{3}\,a = 4r \Rightarrow a = \dfrac{4r}{\sqrt{3}}$
Number of atoms per bcc unit cell $= 8 \times \dfrac{1}{8} + 1 = 2$
Volume occupied by the atoms $= 2 \times \dfrac{4}{3}\pi r^3 = \dfrac{8}{3}\pi r^3$
Volume of the unit cell $= a^3 = \left(\dfrac{4r}{\sqrt{3}}\right)^3 = \dfrac{64r^3}{3\sqrt{3}}$
Packing fraction $= \dfrac{\frac{8}{3}\pi r^3}{\frac{64 r^3}{3\sqrt{3}}} = \dfrac{\sqrt{3}\,\pi}{8} = 0.68$
So 68 percent of the unit cell is occupied by atoms.
Percentage of free space $= 100 - 68 = 32$ percent.