length of the unit cell edge = 400Pm =400x10^-10 cm
 volume of the unit cell  = ( 400 x10^-10cm)³
                                 = 6.4 x 10^-23 cm^-3

as the element A forms a body centred  cubic lattice so no of atom per units cell is 2

z= 2 atoms unit cell

atomic mass of the element  =100 g/mol
density of element is given by

$$\begin{gathered} \mathrm{md}= \frac{\mathrm{M} \mathrm{xZ}}{{\mathrm{a}}^3 \mathrm{xN}{}_{\mathrm{A}}} \\ \mathrm{d}= \frac{100 \mathrm{x}2}{6.4 \mathrm{x} {10}^{-23} \mathrm{x} 6.023 \mathrm{x} 10{ }^{23}} = 51.88 \mathrm{g}/{\mathrm{cm}}^3 \\ \mathrm{volume} \mathrm{of} 10 \mathrm{g} \mathrm{of} \mathrm{A} = \frac{\mathrm{mass}}{\mathrm{Density} } \\ = \frac{10\mathrm{g}}{5.188\mathrm{g} {\mathrm{cm}}^{-3}} \\ = 1.9275 {\mathrm{cm}}^3 \\ \mathrm{number} \mathrm{of} \mathrm{unit} \mathrm{cell} \mathrm{in} 1.9275 {\mathrm{cm}}^3 \mathrm{volume} =\frac{\mathrm{volume} \mathrm{of} \mathrm{the} \mathrm{substance} }{\mathrm{unit} \mathrm{cell} \mathrm{volume} } \\ = 3.0 \mathrm{x}{10}^{22} \mathrm{unit} \mathrm{cell} \end{gathered}$$