I see you already did this in Part II, good! The procedure says to take an average of the iodine molarities calculated from the best titrations. These would be with the volumes 12.40 mL, 12.35 mL, and 12.35 mL of iodine solution. All three of these gave end points that "just changed color." That will finish Part I.
Molarity of the iodine solution (Part I):
M1V1= M2V2
M2= (0.002839M * 0.010L) / 0.01235 L
= 0.002299 M
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