The number of water molecules in gypsum, dead burnt plaster and plaster of Paris, respectively are:

First, we recall the chemical names and formulae of the three calcium sulphate materials mentioned:

We have $$\text{gypsum} = \mathrm{CaSO_4\cdot 2H_2O}.$$

Now, the coefficient of $$\mathrm{H_2O}$$ in the formula tells us exactly how many molecules of water are present in one formula unit. Here the coefficient is $$2$$, so gypsum contains $$2$$ molecules of water of crystallisation.

When gypsum is heated strongly enough to drive out all its water, the product is called dead-burnt plaster. The formula becomes simply $$\mathrm{CaSO_4}$$ with no attached water. So the number of water molecules in dead-burnt plaster is $$0$$.

If, however, gypsum is only partially dehydrated—specifically, if one and a half molecules of water are removed from every two molecules originally present—we obtain plaster of Paris. The well-known formula for plaster of Paris is written as

$$\mathrm{CaSO_4\cdot\frac{1}{2}H_2O}.$$

Here the fraction $$\dfrac{1}{2}$$ in front of $$\mathrm{H_2O}$$ means that in the average crystal lattice there is half a molecule of water per formula unit. Thus plaster of Paris contains $$0.5$$ molecule of water.

Collecting these results together, we have:

$$ \begin{aligned} \text{Gypsum} &:& 2 \text{ water molecules},\\ \text{Dead-burnt plaster} &:& 0 \text{ water molecules},\\ \text{Plaster of Paris} &:& 0.5 \text{ water molecule}. \end{aligned} $$

Looking at the options, the sequence $$2,\;0,\;0.5$$ appears in Option A.

Hence, the correct answer is Option A.