Answer
$\rho=2.59\frac{g}{cm^3}$
Work Step by Step
The radius is .215nm, and the atomic weight is 87.62g\mol.
$\rho=\frac{nA}{{V_{c}}{N_{A}}}$
for FCC, n=4 and $V_{c}=({16r^3\sqrt{2}})$
$N_{A}=6.023\times10^{23} \frac{atom}{mol}$ is Avogadro's number.
$\rho=\frac{4\times87.62}{({16r^3}{\sqrt{2}})\times6.023\times10^{23}}$
$\rho=2.59\frac{g}{cm^3}$
