A metal M readily forms water soluble MSO₄, water insoluble M(OH)₂ and oxide MO which becomes inert on heating. The hydroxide is soluble in NaOH. Then M is:

Given, M reacts to give MSO₄, which is water soluble.
M reacts to give M(OH)₂, which is water insoluble.
M reacts to give MO, which becomes inert on heating.
M(OH)₂ is soluble in NaOH, this indicates that M is Be, since only Be, due to its small size and high ionisation enthalpy, is amphoteric and gives M(OH)₂ which dissolves in NaOH.

$\mathrm{Be}(\mathrm{OH}{)}_2+2\mathrm{NaOH}⟶{\mathrm{Na}}_2{\mathrm{BeO}}_2+2{\mathrm{H}}_2\mathrm{O}$

Other facts given are true for Be and not for others.