The oxidation states of the most electronegative element in the products of the reaction between ${BaO}_2$ and dilute $H_2{SO}_4$ are

The reaction between ${BaO}_2$ and dilute $H_2{SO}_4$ is as follows:

${BaO}_2+H_2{SO}_4\to {BaSO}_4+H_2{O}_2$

From the above reaction it is clear that the most electronegative element in the product is oxygen.

The oxidation state of oxygen in ${BaSO}_4$ is $-2$ and in $H_2{O}_2$ is $-1$.

Hence the correct option is (B).