Answer
$17.7\,g/cm^{3}$
Work Step by Step
Number of atoms in a bcc unit cell: $n=2$
Edge length, $a=\frac{4}{\sqrt 3}r=\frac{4}{\sqrt 3}(141\,pm)=325.6\times10^{-10}\,cm$
Molar mass of tungsten, $M=183.84\,g$
Avogadro's number, $N_{A}=6.022\times10^{23}$
Density= $\frac{n\times M}{a^{3}\times N_{A}}=\frac{2\times183.84\,g}{(325.6\times10^{-10}\,cm)^{3}\times6.022\times10^{23}}$
$=17.7\,g/cm^{3}$
