The oxidation states of the most electronegative element 1n the products of the reaction of ${\mathrm{BaO}}_2$ with dilute ${\mathrm{H}}_2{\mathrm{SO}}_4$ is:

Since oxygen is the most electronegative element the oxidation state of oxygen in ${\mathrm{BaSO}}_4$ is -2 and in ${\mathrm{H}}_2{\mathrm{O}}_2$ is -1.
${\mathrm{BaO}}_2+{\mathrm{H}}_2{\mathrm{SO}}_4\to {\mathrm{BaSO}}_4+{\mathrm{H}}_2{\mathrm{O}}_2$
Hence the oxidation states of the oxygen are -1 and -2.
Hence correct option is B.