What is the oxidation number for the C in C2O42-?

O has an oxidation number of -2 which would give an overall charge of -8 due to oxygen in the molecule [C2O4]2- since there are four O's (4*-2 = -8).  There is also an overall -2 charge on the molecule, meaning that the positive charge due to carbon can only cancel out a charge of -6 due to the oxygen so that the overall molecule has a -2 charge.

Since there are two Carbon atoms, each would have to have a +3 oxidation number so that an overall +6 charge results from carbon, cancelling out only -6 of the -8 charge due to oxygen, and leaving you with a molecule of net charge 2-, exactly what you want.

What seemed to be missing in your thought process was that you gave carbon a negative oxidation number when it must have a positive oxidation number.  Otherwise, you could never balance out the overall charge on [C2O4]2- to be -2 as you would be making it more negative.

It's been a while...

Total charge is 2-.
O charge is -2 (this negative 2 is oxygen's charge) x 4 = -8

So X + -8= -2 (this negative 2 is the total charge, it's what's given)
so X being carbon has to equal 6
6/2 because there are two carbons, an individual carbon has +3.

What's wrong with +3 ? The carbon atom still bond four times, but the central bond is homonuclear, so non-polar.

When this ion reacts, it is oxidised to CO2  where the C oxidation number is +4

So in ethanedioate the carbon is +3, which gets oxidised to carbon dioxide in which carbon is +4. No problem!