In which of the following compounds, the oxidation number of oxygen is $+2$?

Fluorine is the most electronegative element in the periodic table and always exhibits an oxidation state of $-1$ in its compounds.Let's calculate the oxidation state of oxygen ($x$) in Oxygen Difluoride (${\text{OF}}_2$):
$$x+2(\text{Oxidation state of F})=0$$
$$x+2(-1)=0$$
$$x-2=0⟹x=+2$$
In almost all other standard metal oxides, peroxides, and superoxides (like ${\text{Na}}_2\text{O}$ or ${\text{NaO}}_2$), oxygen has a negative oxidation state (such as $-2$ or $-\frac{1}{2}$) because metals are less electronegative than oxygen. Therefore, oxygen has an oxidation state of $+2$ only in ${\text{OF}}_2$.