solution:
[tex]Ta is BCC, a=\frac{4}{\sqrt{3}r},n=2atoms/cell\\
p=16.6gram/cm^3,A=180.9gram/mol\\
P=\frac{A\times n}{N_{a}\times v}=\frac{180.9\frac{gram}{mol}\times 2 \frac{atom}{cell}}{6.023\times10^23\frac{atom}{mol}\times(\frac{4}{\sqrt{3}\times r})^3\frac{cm^3}{cell}}\\
=16.6\frac{gram}{cm^3}\\
r=(\frac{180.9\times2}{16.6\times6.023\times10^23\times(\frac{4}{\sqrt{3}})^3})^\frac{1}{3}\times10^8[/tex]