The correct order of the solubility of alkaline-earth metal sulphates in water is:

First, let us recall a key idea from general inorganic chemistry. When an ionic solid such as a sulphate $$M^{2+}SO_4^{2-}$$ (where $$M^{2+}$$ is an alkaline-earth metal ion) dissolves in water, two enthalpy terms dominate the energetics:

$$\Delta H_{\text{solution}} \;=\; \Delta H_{\text{lattice}} \;+\; \Delta H_{\text{hydration}}$$

Here,

$$\Delta H_{\text{lattice}}$$ is the lattice enthalpy, the energy required to pull the $$M^{2+}$$ and $$SO_4^{2-}$$ ions apart from the crystal, so it is always positive.

$$\Delta H_{\text{hydration}}$$ is the hydration enthalpy, the energy released when those free ions are surrounded by water molecules, so it is always negative.

An ionic salt will be readily soluble when the magnitude of the (negative) hydration enthalpy outweighs the (positive) lattice enthalpy, making the overall $$\Delta H_{\text{solution}}$$ close to zero or negative. Therefore, for a series of similar salts, greater (i.e., more negative) hydration enthalpy leads to greater solubility, provided lattice energies do not change dramatically.

Now we examine how each of these enthalpies changes down Group 2, from magnesium to barium.

As we move down the group

$$Mg^{2+} \;\rightarrow\; Ca^{2+} \;\rightarrow\; Sr^{2+} \;\rightarrow\; Ba^{2+},$$

the ionic radius increases. A larger ion has a lower charge-to-radius ratio, so water molecules are held less strongly. Hence, the magnitude of the hydration enthalpy decreases sharply:

$$|\Delta H_{\text{hydration}}(Mg^{2+})| \;>\; |\Delta H_{\text{hydration}}(Ca^{2+})| \;>\; |\Delta H_{\text{hydration}}(Sr^{2+})| \;>\; |\Delta H_{\text{hydration}}(Ba^{2+})|.$$

Lattice enthalpy also declines down the group because larger cations produce a smaller electrostatic attraction within the crystal. However, this fall is relatively gradual for sulphates.

The crucial point is that the decrease in the magnitude of hydration enthalpy down the group is much greater than the decrease in lattice enthalpy. Mathematically, if we write the change from one metal to the next as

$$\Delta H_{\text{solution,new}} - \Delta H_{\text{solution,old}} \;=\; (\Delta H_{\text{lattice,new}} - \Delta H_{\text{lattice,old}}) + (\Delta H_{\text{hydration,new}} - \Delta H_{\text{hydration,old}}),$$

the second bracketed term (coming from hydration) is numerically larger and positive (because the negative value becomes less negative), so $$\Delta H_{\text{solution}}$$ becomes less negative or even positive. Consequently the salt becomes progressively less soluble.

Therefore, for the alkaline-earth metal sulphates the solubility trend is

$$MgSO_4 \;>\; CaSO_4 \;>\; SrSO_4 \;>\; BaSO_4.$$

Translating that chemical order to the symbols of the metals alone, we obtain

$$Mg \;>\; Ca \;>\; Sr \;>\; Ba.$$

This exactly matches Option A in the list:

Mg > Ca > Sr > Ba

Hence, the correct answer is Option A.