Answer
$r=0.1432nm$
Work Step by Step
Given ${\rho}=16.6\frac{g}{cm^3}$, Atomic weight=180.9g\mol, and
$\rho=\frac{nA}{{V_{c}}{N_{A}}}$
For BCC, n=2 and $V_{c}=(\frac{4r}{\sqrt{3}})^3$
$N_{A}=6.023\times10^{23} \frac{atom}{mol}$ is Avogadro's number
$16.6=\frac{2\times180.9}{(\frac{4r}{\sqrt{3}})^3\times6.023\times10^{23}}$
$r^3=\frac{2\times3\sqrt{3}\times180.9}{({16.6\times16}{\sqrt{2}})\times6.023\times10^{23}}$
$r=0.1432nm$
