From eight corner atoms one atoms (X) contributes to one unit cell. From six face centres, three atoms (Y) contributes to one unit cell. So, the formula of the compound is $XY_3$.

As we know that, $\rho =\large\frac{n\times M_m}{N_A\times a^3}$ , here n = 1

Molar mass of $XY_3$

$M_m = 60 + 3 × 90 = 330$ gm

$\rho=\large\frac{1\times 330}{6.023\times 10^{23}\times (5\times 10^{-8})^3}$$gm/cm^3$

a = $5A^{\large\circ} = 5 × 10^{–8} $cm

$\Rightarrow \large\frac{330}{6.023\times 10^{23}\times 125\times 10^{-24}}$$gm/cm^3=4.38gm/cm^3$

Hence (B) is the correct answer.