As the size of central atom, (order of size $\mathrm{O}>\mathrm{S}>\mathrm{Se}>\mathrm{Te}$ ) increases, the $\mathrm{H}-A$ (where $A=$ central atom) bond length increases. Thus, the $\mathrm{H}-\mathrm{A}$ bond dissociation energy decreases. Consequently $\mathrm{H}_2$ Te gives a proton $\left(\mathrm{H}^{+}\right)$more readily. Hence, it is most acidic among the given.
$\mathrm{H}_2 \mathrm{Te}+\mathrm{H}_2 \mathrm{O} \longrightarrow \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{HTe}^{-}$