$d=\large\frac{z\times M}{a^3\times N_A}$

No. of atoms in fcc lattice z=4

$d=\large\frac{4\times 63.5mol^{-1}}{(3.61\times 10^{-8}cm)^3\times (6.02\times 10^{23}mol^{-1})}$

Atomic mass of unit cell =$\large\frac{4\times 63.5}{6.02\times 10^{23}}$g

$\Rightarrow 4.22\times 10^{-22}$g

Density =$\large\frac{\text{Mass of unit cell}}{\text{Volume of unit cell}}$

$\Rightarrow \large\frac{4.22\times 10^{-22}}{47.4\times 10^{-24}cm^2}$

$\Rightarrow 8.9gcm^{-3}$

Which is in close agreement with the measured value.

Density = Mass of unit cell/Volume of unit cell

$8.93 = (4 \times 63.5)/(a^3 \times 6.02 \times 10^{23}).$

$a^ 3 = 47.24 x 10^{-24 }$

$\qquad= 360\;pm$