How many electrons are transferred in 3I2(s) + 2MnO2(s) + 8OH-(aq) --> 6I-(aq) + 2MnO4-(aq) + 4H2O(l)?

Equation doesn't make too much sense as written. However, we do not do homework for you. Show us your work.

The oxidation state of iodine goes from 0 to -1. Since there are 6 iodine ions, the amount of electrons transferred is -1 x 6= -6.
The oxidation state of 2MnO2 goes from 0 to -1. Because there are two of these ions in the products, 2 x -1 = -2. Therefore there are 2 electrons that have been transferred.
Total electrons gained by reactants= -8
The oxidation state of OH is -1. During reaction, oxidation state changes to 0. Since there are 8 OH ions, total oxidation state= 8 x 1 = +8

Thus, there is a total amount of 8 electrons that have been transferred (from reactants to products) during this reaction.

I have worked this out, however I don't know if it is correct or not.
Your help would be greatly appreciated.

When I got my degree in 1963 we used."valence " as the electrons. It's been a while since I balanced equations but your math appears correct -at least in the newer terminology.