We know that
$Z = \large\frac{a^3\times d\times N_A}{M}$
$\therefore d = \large\frac{ZM}{a^3N_A}$
Z = 4 for fcc unit cell ;
$M = 63.5 g\; mol^{-1}$
$a = 361.6\times 10^{-10}cm$
$N_A = 6.023\times10^{23}$
Putting these values in (i) we get
$d = \large\frac{4\times63.5}{(361.6\times10^{-10})^3\times 6.023\times10^{23}}$
$=8.94 g\;cm^{-3}$
Hence answer is (B)