Answer
$r=0.139nm$
Work Step by Step
Given ${\rho}=12.0\frac{g}{cm^3}$, atomic weight=106.4g\mol, and
$\rho=\frac{nA}{{V_{c}}{N_{A}}}$
For FCC, n=4 and $V_{c}=({16r^3\sqrt{2}})$
$N_{A}=6.023\times10^{23} \frac{atom}{mol}$ is Avogadro's number
$\rho=\frac{4\times106.4}{({16r^3}{\sqrt{2}})\times6.023\times10^{23}}$
$r^3=\frac{4\times106.4}{({16\times12}{\sqrt{2}})\times6.023\times10^{23}}$
$r=.139 nm$
