Answer : $2.586gcm^{-3}$
The cubical closest-packed structure has a face-centred cubic unit cell.In the latter atoms touch along the face diagonal of the cube.Hence
$\sqrt 2a=4r=4(215 pm)=860$pm or $\;\;a=\large\frac{860pm}{\sqrt 2}$$=608.2$pm
Now using the expression
$\rho=\large\frac{N}{a^3}\big(\frac{M}{N_A}\big)$
We get $\rho=\large\frac{4}{(608.2\times 10^{-12}m)^3}\big(\frac{87.6gmol^{-1}}{6.023\times 10^{23}mol^{-1}}\big)$$=2.586\times 10^6gm^{-3}\equiv 2.586gcm^{-3}$