In the following reaction :

$\mathrm{Cr}(\mathrm{OH}{)}_3+{\mathrm{OH}}^{-}+{\mathrm{IO}}_3^{-}⟶{\mathrm{CrO}}_4^{2-}+{\mathrm{H}}_2\mathrm{O}+{\mathrm{I}}^{-}$

Since oxidation number of I changes from $+5$ to $-$. ${\mathrm{IO}}_3^{-}$is the oxidizing agent. While the oxidation number of chromium changes from  $+3\text{ to }+6\mathrm{Cr}(\mathrm{OH}{)}_{3}9$ is oxidized.

The redox reaction is shown below,

$\mathrm{Cr}(\mathrm{OH}{)}_3+(\mathrm{OH}{)}^{-}+{\mathrm{IO}}_3^{-}\to {\mathrm{CrO}}_4^{2-}+{\mathrm{H}}_2\mathrm{O}+{\mathrm{I}}^{-}$

Hence, options A and B are correct.