A metal M crystallizes into two lattices :- face centred cubic...

The density of the crystal lattice can be calculated as follows,

$d = \dfrac{Z \times M}{{N}_{A}{a}^{3}}$

$Z =$effective number of atoms in unit cell(For BCC=2 and FCC=4)

$M =$Molar mass

${N}_{A}$=Avogadro's number

$a =$Edge length

$\dfrac{{d}_{FCC}}{{d}_{BCC}} = \dfrac{\dfrac{4 \times {M}_{w}}{{N}_{A} \times ( 2{)}^{3}}}{\dfrac{2 \times {M}_{w}}{{N}_{A} \times ( 2 . 5{)}^{3}}} = 3 . 90$