An element forms a body centered cubic (bcc) lattice with edge length of 300 pm, If the density of the element is $7.2gc{m}^{-3}$,the number of atoms present in $324g$ of it approximately is

Volume of bcc unit cell = 300 pm

$={(300\times {10}^{-10}cm)}^3=27\times {10}^{-24}c{m}^3$

Volume of 324 g of the element =$\frac{Mass}{Density}$

$=\frac{324g}{7.2gc{m}^{-3}}=45c{m}^3$

For a bcc structure, number of atoms per unit cell = 2

So, number of  atoms = $\frac{Totalvolume}{volumeofaunitcell}$

$=\frac{2\times 45}{27}\times {10}^{24}=333\times {10}^{24}$