A metal $\text{'}\mathrm{M}\text{'}$ readily gives ${\mathrm{MSO}}_4,$ which is soluble in water. It forms its oxide $\mathrm{M}0$ which is amphoteric. It forms an insoluble hydroxide $\mathrm{M}(\mathrm{OH}{)}_2,$ which is soluble in $\mathrm{NaOH}$ solution. The $\mathrm{M}$ is

According to the question, the metals given in options belong alkaline earth metals. 

The solubility of sulphate of alkaline earth metals decreases on the going down the group because Be to Ra, the magnitude of hydration enthalpy decreases significantly, due to increase in the atomic size of alkaline earth metals, bringing about a decrease in solubility of their sulphates. 

In case of alkaline earth metals, Be, Mg or Ca forms monoxide whereas Ba and Sr forms peroxide. In case of BeO is amphoteric while oxides of other elements are basic in nature. 

The hydroxide of Mg, Ca, Sr & Ba are base and their basic strength increases from Mg tot Ba. The solubility of the hydroxides of alkaline earth metals in water increases with increase in atomic number, ${\text{Be(OH)}}_{\text{2}}{\text{ \& Mg(OH)}}_{\text{2}}$ Are almost insoluble in water, $\mathrm{Ca}{\left(\mathrm{OH}\right)}_2$ Is sparingly soluble while $\mathrm{Sr}{\left(\mathrm{OH}\right)}_2 \& \mathrm{Ba}{\left(\mathrm{OH}\right)}_2$ Are increasingly more soluble. Due to the amphoteric nature of BeO it is soluble in NaOH.

All these properties, given in the question belong to be, so option (a) is correct.