The oxidation state of Copper here will be 2+ because of the sulfate moiety (2-). The correct answer therefore should be C, Cu to Cu +2e-. The 2+ oxidation state results in 2 free electrons on the right side of the equation, a reduction.
The oxidation number of Cu in CuSO₄ is +2 and 0 in Cu, the free elemental form. As the oxidation of Cu decreases from +2 to 0, we are looking at the reduction half-reaction for the given redox reaction:
