$(a)\;3 \;cm\qquad(b)\;4\;cm\qquad(c)\;5\;cm\qquad(d)\;6\;cm$

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We know that ,

$\rho = \large\frac{N}{a^3} (\large\frac{M}{N_A})$

Where

$\rho = density = 2.165 gcm^{-3}$

M = molar mass = 58.5

$N_A = Avogadro's\; number = 6.023\times 10^{23}$

N = number of formula unit per unit cell = 4 (for fcc)

$a^3 = \large\frac{N}{\rho}(\large\frac{M}{N_A})$

$= \large\frac{4}{2.165}[\large\frac{58.5}{6.023\times10^23}]$

$=1.794\times10^{-22}$

$a = 5.64\times10^{-8}cm$

Molar Volume = $\large\frac{Molar\;mass}{Density}$

$=\large\frac{58.8}{2.165}$

Edge length (a) = $[\large\frac{58.8}{2.165}]^{\large\frac{1}{3}}$

$\;\;\;\;\;\;\;\;\;\;\;\; = 3 cm$

Hence answer is (A)

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