How many water molecules are present in $54$ gram of water?

We get $1.81\times {10}^{24}$ water molecules...

Explanation:

We assess the molar quantity of water in the usual way...

$\text{Number of moles}=\frac{\text{mass}}{\text{molar mass}}$

$\equiv \frac{54\cdot g}{18.01\cdot g\cdot mo{l}^{-1}}=3.00\cdot mol$

But the mol is like the dozen or the gross, it specifies a NUMBER, here $N_A\equiv 6.022\times {10}^{23}\cdot mol$. Why do we use such an absurdly large number? Because $N_A\cdot 1H$⋅atoms have a MASS of $1\cdot g$ precisely...and thusthe mole is the link between the micro-world of atoms, and molecules, to the macro world of grams and litres....

And so (finally!) we take the product...

$\text{Number of water molecules}=\text{molar quantity}\times \text{Avogadro's number.}$

$=\underset{\text{a number as required..}}{\underset{}{3.00\cdot mol\times 6.022\times {10}^{23}\cdot mo{l}^{-1}}}=??\cdot \text{water molecules}$