Answer
$V_{c}=1.06\times{10}^{-28}{m^3}$
Work Step by Step
Given Atomic weight=47.87g\mol $\rho=4.51$ 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
$4.51=\frac{6\times47.87}{(V_{c})\times6.023\times10^{23}}$
$V_{c}=1.06\times{10}^{-28}{m^3}$
