Option 1:  ${\mathrm{Mn}}^{+2}\left[\mathrm{Ar}\right]3 {\mathrm{d}}^5,{\mathrm{Cr}}^{+}\left[\mathrm{Ar}\right]3 {\mathrm{d}}^5$

Cr is expected to have the highest second ionization enthalpy as after removal of 4s electrons, the electrons will be removed from d-orbital. Cr+ have stable half-filled d-orbital with maximum spin multiplicity. 

Option $-2 {\mathrm{Ni}}^{+2}\left[\mathrm{Ar}\right]3 {\mathrm{d}}^8,{\mathrm{Cu}}^{+}\left[\mathrm{Ar}\right]3 {\mathrm{d}}^{10}$

Option $-3 {\mathrm{Fe}}^{+2}\left[\mathrm{Ar}\right]3 {\mathrm{d}}^6,{\mathrm{Co}}^{+}\left[\mathrm{Ar}\right]3 {\mathrm{d}}^{7}4 {\mathrm{s}}^1$

Option $-4 {\mathrm{V}}^{+2}\left[\mathrm{Ar}\right]3 {\mathrm{d}}^3,{\mathrm{Cr}}^{+}\left[\mathrm{Ar}\right]3 {\mathrm{d}}^5$