Answer
BCC
Work Step by Step
r=0.143nm, Atomic weigh=92.97g\mol, and $\rho=8.57\frac{g}{cm^3}$
$\rho=\frac{nA}{{V_{c}}{N_{A}}}$
We consider each crystal structure:
For FCC, n=4 and $V_{c}=({16r^3\sqrt{2}})$
$N_{A}=6.023\times10^{23} \frac{atom}{mol}$ is Avogadro's number.
$\rho=\frac{4\times92.97}{({16r^3}{\sqrt{2}})\times6.023\times10^{23}}$
$\rho(fcc)=9.33\frac{g}{cm^3}$
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.
$\rho=\frac{2\times92.97}{(\frac{4r}{\sqrt{3}})^3\times6.023\times10^{23}}$
$\rho(bcc)8.567\frac{g}{cm^3}$
Since this is the density for BCC, it follows that the sample is BCC.
