From the given information
The number of moles of Iodine produced in the reaction is calculated by using the titration of the Iodine with Hypo.
In the given reaction ,
The number of moles of Hypo produced = Molarity * Volume(in L)
= 0.10 M * 0.0656 L
= 0.00656 moles
According to the titration reaction
I2 + 2Na2SO3 --> 2NaI + Na2S2O6
1 mole of I2 reacted with the 2 moles of the hypo.Thus in the present case we have 0.00656 moles of the hypo.Thus the total number of moles of the Iodine produced.
= 0.00656 mole Hypo * (1mole of I2 / 2 moles of Hypo)
= 0.00328 moles of I2 is produced.