The products formed when an aqueous solution of $\mathrm{NaBr}$ is electrolyzed in a cell having inert electrodes are :

The products formed when an aqueous solution of $\mathrm{NaBr}$ is electrolyzed in a cell having inert electrodes are ${\mathrm{H}}_2,{ \mathrm{Br}}_2$ and $\mathrm{NaOH}$.

$\begin{array}{l}\mathrm{NaBr}\rightleftharpoons {\mathrm{Na}}^{+}+{\mathrm{Br}}^{-}\\ 2{\mathrm{H}}_2\mathrm{O}+2\mathrm{e}\to {\mathrm{H}}_2+2{\mathrm{OH}}^{-}\end{array}$

${\mathrm{Na}}^{+}+{\mathrm{OH}}^{-}\to \mathrm{NaOH}$ at cathode

${\mathrm{Br}}^{-}\to \mathrm{Br}+{\mathrm{e}}^{-}$ 

$\mathrm{Br}+\mathrm{Br}\to {\mathrm{Br}}_2$ at anode

So the products are ${\mathrm{H}}_2$ and $\mathrm{NaOH}$ (at cathode) and ${\mathrm{Br}}_2$ (at anode).