Heat of neutralization of ${\mathrm{H}}_2{\mathrm{SO}}_4$ with $\mathrm{NaOH}$ is $-x\mathrm{KJ}$. Then the heat of ionisation of water is

Chemical reaction

${\mathrm{H}}_2{\mathrm{SO}}_4⟶2{\mathrm{H}}^{+}+{\mathrm{SO}}_4^{2-}$

$2{\mathrm{H}}^{+}+2{\mathrm{OH}}^{-}\rightleftharpoons 2{\mathrm{H}}_2\mathrm{O}; \Delta {\mathrm{H}}^{-}=-x{ \mathrm{kJ}}^2$

Ionization of Water is the reverse of heat of neutralization \  the value of heat will be equal but in reverse Sign.

The expression of ionization of Water

$2{\mathrm{H}}_2\mathrm{O}\rightleftharpoons 2{\mathrm{H}}^{+}+2{\mathrm{OH}}^{-}; \Delta \mathrm{H}=+x \mathrm{kJ}$

Hence the correct answer is (B) $x\mathrm{KJ}$.