Consider the change in oxidation state of Bromine corresponding to...

Calculate ${E}_{cell}^{o}$ corresponding to each compound undergoing disproportionation reaction. The reaction for which ${E}_{cell}^{o}$ comes out +ve is spontaneous.

$\text{HBrO} \to \text{B}{\text{r}}_{2}{\text{E}}^{\text{o}} = 1.595, \text{S}\text{R}\text{P}\left(\text{Cathode}\right)$
$2\text{HBrO} \to \text{B}{\text{r}}_{2} + \text{B}\text{r}{\text{O}}_{3}^{-}$
${\text{E}}_{\text{c}\text{e}\text{l}\text{l}}^{\text{o}} =$ SRP (cathode) - SRP (Anode)

$= 1.595 - 1.5$

$= 0 .095 V$

${\text{E}}_{\text{cell}}^{\text{o}} > 0 ⇒ \Delta{\text{G}}^{o} < 0$ [spontaneous]