Answer
$0.533\,g/cm^{3}$
Work Step by Step
Number of atoms in a bcc unit cell: $n=2$
Edge length, $a=351\,pm=351\times10^{-10}\,cm$
Molar mass of Li, $M=6.941\,g$
Avogadro's number: $N_{A}=6.022\times10^{23}$
Density= $\frac{n\times M}{a^{3}\times N_{A}}=\frac{2\times6.941\,g}{(351\times10^{-10}\,cm)^{3}\times6.022\times10^{23}}$
$=0.533\,g/cm^{3}$
