$\begin{array}{1 1}2.586gcm^{-3}\\1.586gcm^{-3}\\3.586gcm^{-3}\\0.586gcm^{-3}\end{array} $

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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}$

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