This question is testing whether you understand whether electrons are removed from 4s or 3d first.
The choices are all transition metals with d shells. A d subshell has 5 orbitals. Then each orbital can carry 2 electrons. According to the Hund's rule, every orbital in a subshell is singly occupied with one electron before any one orbital is doubly occupied.
A. 3 unpaired electrons. Cr is [Ar] 4s^1 3d^5. Then removing 1 electron from the 4s subshell and 2 electrons from 3d subshell, you should get Cr3+ : [Ar]3d^3 --
B. Mn2+: [Ar]3d^5-- 5 unpaired electrons --You have to remove the electrons from the highest shell number first, which is 4s
C. Cu+: no unpaired electrons! Cu: is 4s^1 3d^10. So Cu+ will be: [Ar]3d^10. You removed the electron from the 4s shell first.
D. Ni2+: 2 unpaired electrons. Remember, [Ar] 3d8 4s2 for Ni and [Ar] 3d8 for Ni+2 because the electrons are removed from the 4sh shell first.