Density = Mass/Volume
Mass of n molecules of $H_2O$ in hexagonal packing = $\large\frac{18n}{N_{\circ}} = \large\frac{18n}{6.023\times 10^{23}}$
Given Volume = $132\times 10^{-24}cm^3$
Density = $0.92g/cm^3$
Substituting the value we get
$0.92 = \large\frac{18n}{ 6.023\times 10^{23}\times 132\times 10^{-24}}$
$\therefore n = \large\frac{0.92 \times 6.023 \times 10^{23} \times 132\times 10^{-24}}{18}$
$\therefore n = 4$