The answer is MnO42- but I dont really understand why.
You just have to count the d-Electrons. Mn(VI) in MnO
42- owns only one of them.
I guess that the ligands in the rest of the complex ions also donate electrons to the charged transition ions?
No that is not the case. Ti(IV) in TiCl4 has d0 configuration and for the Co(III) complex the six d-electrons are paired due to the low-spin configuration.
It might be worth noting that Fe(III) in complex C has d5 high-spin configuration and, thus, five unpaired electrons.