Answer
$1.00\,g/cm^{3}$
Work Step by Step
The bcc unit cell contains $8\times\frac{1}{8}+1=2$ Na atoms.
Mass in grams of Na per unit cell=
$\frac{\text{2 Na atoms}}{\text{unit cell}}\times\frac{1\,mol\,Na}{6.022\times10^{23}\,Na\,atoms}\times\frac{23.0\,g\,Na}{1\,mol\,Na}$
$=7.638658\times10^{-23}\, g$
Volume of the unit cell= $(4.24\times10^{-8}\, cm) ^{3}$
$=7.6225\times10^{-23}\, cm^{3}$
$\text{Density}= \frac{\text{Mass}}{\text{Volume}}$
$=\frac{7.638658\times10^{-23}\, g}{7.6225\times10^{-23}\, cm^{3}}=1.00\,g/cm^{3}$
