The average O.N. of S in $\mathrm{S_2O_3^{2-}}$ is +2 while in $\mathrm{S_4O_6^{2-}}$ it is +2.5. The O.N. of S in $\mathrm{SO_4^{2-}}$ is +6. Since $\mathrm{Br_2}$ is a stronger oxidising agent, it oxidises S of $\mathrm{S_2O_3^{2-}}$ to a higher oxidation state of +6 and hence forms $\mathrm{SO_4^{2-}}$ ion. $\mathrm{I_2}$, however, being a weaker oxidising agent oxidises S of $\mathrm{S_2O_3^{2-}}$ ion to a lower oxidation state of + 2.5 in $\mathrm{S_4O_6^{2-}}$ ion.