Atomic weight of metal is 55.85. Avogadro number is 6.023×1023/gm mole.

Let $\mathrm{a}=$edge of unit cell

=lattice parameter of a cubic cell

$M=$ mass of element per mole

= atomic weight of the element

$\rho=$ density of the metal

$\mathrm{N_A}=$ Avogadro's number

n= number of atoms per unit cell

Thus

$\mathrm{a=2.9}\dot{A}=2.9\times10^{-8} cm$

$\mathrm{M=55.85}$

$\mathrm{\rho=7.87 \space gm/cc}$

$\mathrm{N_A=6.023\times 10^{23}}$

Number of atoms per unit cell,n:

Using the relation

$\mathrm{\rho\space a^{3} =\dfrac{nM}{N_A}}$

or $\mathrm{n=\dfrac{\rho\space a^3N_A}{M}}$

$\mathrm{=\dfrac{7.87\times(2.9\times 10^{-8})^3\times 6.0238\times 10^{23}}{55.85}=2}$