$(a)\;2\qquad(b)\;3\qquad(c)\;4\qquad(d)\;5$

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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)

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