The equivalent weight of Na2S2O3 in the reaction is
2Na2S2O3+ I2gives Na2S4O6 + 2NaI

The equivalent weight of oxidising or reducing agent is = molecular weight / number of electron per mole transfer during the reaction.
In above mentioned reaction there is transfer of 2 electrons per 2 moles of Na₂S₂O₃ as follows :
2S₂O₃^2- → S₄O₆^2- + 2e-
So the equivalent weight of Na₂S₂O₃ is same as molecular weight = 158.11 g mol^-1​.