If $E^{\circ }\left({\mathrm{BrO}}^{-},{\mathrm{BrO}}_3^{-},{\mathrm{OH}}^{-}\mid \mathrm{Pt}\right)=0.54\mathrm{V}$ and $E^{\circ }\left({\mathrm{BrO}}^{-},{\mathrm{Br}}_2,{\mathrm{OH}}^{-}\mid \mathrm{Pt}\right)=0.45\mathrm{V}$ the $E^{\circ }$ value of the half-cell ${\mathrm{BrO}}_3^{-},{\mathrm{Br}}_2,{\mathrm{OH}}^{-}\mid \mathrm{Pt}$ will be 

$\begin{array}{l}{\mathrm{BrO}}_3^{-}+2{\mathrm{H}}_2\mathrm{O}+4{\mathrm{e}}^{-}\to {\mathrm{BrO}}^{-}+4{\mathrm{OH}}^{-}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{E}^{\circ }=0.54\mathrm{V}\\ {\mathrm{BrO}}^{-}+{\mathrm{H}}_2\mathrm{O}+{\mathrm{e}}^{-}\to \frac{1}{2}{\mathrm{Br}}_2+2{\mathrm{OH}}^{-}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{E}^{\circ }=0.45\mathrm{V}\end{array}$

For the given half-cell, the reaction is 

${\mathrm{BrO}}_3^{-}+3{\mathrm{H}}_2\mathrm{O}+5{\mathrm{e}}^{-}\to \frac{1}{2}{\mathrm{Br}}_2+6{\mathrm{OH}}^{-}{E}^{\circ }=?$

This reaction is obtained by adding the first two reactions 

Hence, $-5F{E}^{\circ }=-4F\left(0.54\mathrm{V}\right)-1F\left(0.45\mathrm{V}\right)$  $E^{\circ }=0.52\mathrm{V}$