The length of the unit cell edge of a bcc lattice metal is 350 pm. Thus, volume of atoms in one mole of the metal is

In a bcc structure, the atoms touch each other along the body diagonal. 

Radius of the atom = r

${\mathrm{AB}}^2={\mathrm{BC}}^2+{\mathrm{AC}}^2$

${\mathrm{BC}}^2={\mathrm{BD}}^2+{\mathrm{CD}}^2={\mathrm{a}}^2+{\mathrm{a}}^2=2{\mathrm{a}}^2$

$\therefore {\mathrm{AB}}^2=2{\mathrm{a}}^2+{\mathrm{a}}^2=3{\mathrm{a}}^2$

$\therefore \mathrm{AB}=\sqrt{3}\mathrm{a}$

But $\mathrm{AB}=4\mathrm{r}$

$\therefore 4\mathrm{r}=\sqrt{3}\mathrm{a}\Rightarrow \mathrm{r}=\sqrt{3}\mathrm{a}/4$

$\mathrm{r}=\sqrt{3}\times \frac{350}{4}\mathrm{pm}$

$=\sqrt{3}\times \frac{350}{4}\times {10}^{-12}\mathrm{m}$

$=\sqrt{3}\times \frac{350}{4}\times {10}^{-10}\mathrm{cm}$

$=151.55\times {10}^{-10}cm$

Volume of one atom $\frac{4}{3}{\mathrm{\pi r}}^3=\frac{4}{3}\mathrm{\pi }{(154.14\times {10}^{-10})}^3{\mathrm{cm}}^3$

                                                       $=1.45\times {10}^{-23}{\mathrm{cm}}^3$

Volume of ${\mathrm{N}}_0$ atoms is one mole of the metal 

                                                   $=6.02\times {10}^{23}\times 1.45\times {10}^{-23}{\mathrm{cm}}^3=8.733 {\mathrm{cm}}^3$