Solution :
No of atoms per unit cell for $bcc (z) =2$
No of atoms in $9.2\;g$ of $Na$
$\qquad= \large\frac{9.2\;g}{23\;g\;mol^{-1}}$$ \times 6.022 \times 10^{23} atoms mol^{-1}$
$\qquad= 2.4088 \times 10^{23}atoms$
No of unit cells $= \large\frac{2.4088 \times 10^{23}\;atoms}{2\;atoms\;unit\;cell^{-1}}$
$\qquad= 1.2044 \times 10^{3} unit \;cell$