I am currently having this problem on reaction kinetics / titration stochiometry

Reaction: H2O2 + 2H+ + 2I- -> I2 + 2H2O

In an experiment, 25.0 cm3 of a 1.5 mol dm-3 hydrogen peroxide, 25.0cm3 of a 1.5 mol dm-3 of sulfuric acid and 50.0 cm3 of a 0.04 mol dm-3 potassium iodide were mixed. 25.0cm3 of the reaction mixture was pipetted out at regular time intervals, quenched and titrated with 0.0200 mol dm-3 sodium thiosulfate

Calculate the volume of sodium thiosulfate which would be needed to react with the 25.0 cm3 sample when the reaction reached completion?

So I calculate the mol first and found that mol of I- = 50/1000x0.04 = 2x10^-3 mol so mol of I2 = 10^-3 mol?

And I am able to write the next reaction which is :

I2 + 2S2O3 2- -> 2I- + S4O6 2-

So im confused and dont know how to calculate the volume of sodium thiosulfate (S2O3 2-)?

Here was my solution but the answer i got was wrong:

mol of S2O3 2- = 2 x mol I2 = 2x10^-3 mol => Volume of S2O3 = mol of S2O3 / conc. of S2O3 = 2 x 10^-3 / 0.02 = 0.1dm3?