Volume of unit cell V = Area of $base\times height$
$=2\times\large\frac{\sqrt3}{4}a^2\times c$
$\large\frac{\sqrt3}{2}\times(4.53\times10^{-8})^2\times7.4\times10^{-8}$
$=13.17\times10^{-23}cm^3$
Mass of unit cell = $V\times\rho$
$=13.17\times10^{-23}\times0.92$
$=1.21\times10^{-22}g$
Let there are n effective water molecules in the given unit cell. Then n times the mass of each water molecule would be the mass of unit cell.
$\therefore 1.21\times10^{-22} = n\times\large\frac{18}{6.023\times10^{23}}$
$\therefore n =4$
Hence answer is (c)