It is given that density of the element, d = 2.7 x 10^3 kg m-3
Molar mass, M = 2.7 x 10^-2 kg mol^-1
Edge length, a= 405 pm = 405 x 10^-12 m = 4.05 x 10^-10 m
It is known that, Avogadro's number, NA= 6.022 x 10^23 mol^-1
d = (z.M)/(a^3.NA)
z = (d.a^3 NA)/M
= [2.7 x 10^3 kg m^-3 x (4.05 x 10^-10 m)^3 x 6.022 x 10^23 mol^-1]/2.7 x 10^-2 kg mol^-1
= 4.004
= 4
Applying the relation, This implies that four atoms of the element are present per unit cell.