We know that ,
$\rho =\large\frac{N}{a^3}(\frac{M}{NA})\rho=\large\frac{N}{a3(MNA)\rho} = \large\frac{N}{a^3} (\large\frac{M}{N_A})$
Where
$ \rho =density=2.165\;g\;cm^{−3} \rho =density=2.165gcm−3\rho = density = 2.165 gcm^{-3}$
M = molar mass = 58.5
$N_A$=Avogadro′snumber$=6.023 \times 10^{23}N_A$=Avogadro′snumber$=6.023 \times 10^{23}N_A = Avogadro's\; number = 6.023\times 10^{23}$
N = number of formula unit per unit cell = 4 (for fcc)
$a^3=\large\frac{N}{\rho} \frac{M}{N_A}a^3=\frac{N}{\rho}(MNA)a^3 = \large\frac{N}{\rho}(\large\frac{M}{N_A})$
$\qquad=1.794 \times 10^{−22}=1.794×10−22=1.794\times10^{-22}$
$\qquad = 5.64\times10^{-8}cm$
Molar Volume =$\large\frac{Molar\;mass}{Density}$
$\large\frac{58.8}{2.165}$
3 cm
Hence answer is (A)