$2{S_2}O_3^{2-}$ $..$ $\longrightarrow$ $S_4O_6^{2-}$ $..$
(Tetrathionate is produced)
The oxidation nuber of (S) varies from 2 to $\frac{5}{2}$,
therefore,
the oxidation number $2\cdot 2\cdot 2 = 8$
of all the $S$ atoms of $2S_2O_3^{2-}$
goes up to the oxidation number
$4\cdot \frac{5}{2} = 10$
of all the $S$ atoms of $S_4O_6^{2-}$,
so $2e^-$ are lost:
$2S_2O_3^{2-}$ $-$ $2e^-$ $\longrightarrow$ $S_4O_6^{2-}$
Now the complete equation:
$2S_2O_3^{2-}$ $-$ $2e^-$ $\longrightarrow$ $S_4O_6^{2-}$
$I_2$ $+$ $2e^-$ $\longrightarrow$ $2I^-$
$2S_2O_3^{2-}$ $+$ $I_2$ $\longrightarrow$ $S_4O_6^{2-}$ $+$ $2I^-$