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

In bcc unit cell, the number of atoms = 2
Thus, volume of atoms in unit cell $\left(v\right)=2\times \frac{4}{3}\pi r^3$
For bcc structure$\left(r\right)=\frac{\sqrt{3}}{4}a$
$\left(V\right)=2\times \frac{4}{3}\pi {\left(\frac{\sqrt{3}}{4}a\right)}^3=\frac{\sqrt{3}}{8}\pi a^3$
Volume of unit cell $\left(V\right)=a^3$
Percentage of volume occupied by unit cell
$\frac{\text{Volume of the atoms in unit cell}}{\text{Volume of unit cell}}$
$=\frac{\frac{\sqrt{3}}{8}\pi a^3}{a^3}\times 100=\frac{\sqrt{3}}{8}\pi \times 100=68\%$
Hence, the free space in bcc unit cell $=100-68$
$=32\%$