Answer
$V_{c}=1.39\times{10}^{-22}{cm^3}$
Work Step by Step
Atomic weight=24.3g\mol, $\rho=1.74$ and
$\rho=\frac{nA}{{V_{c}}{N_{A}}}$
For HCP, n=6.
$N_{A}=6.023\times10^{23} \frac{atom}{mol}$ is Avogadro's number.
$1.74=\frac{6\times24.3}{(V_{c})\times6.023\times10^{23}}$
$V_{c}=1.39\times{10}^{-22}{cm^3}$
