The difference in the number of unpaired electrons of a metal ion in its high-spin and low-spin octahedral complexes is two. The metal ion is:

The metal ion in question is most likely to be Mn3+. Here's why:

In a high-spin complex, the energy gap between the lower and higher energy sets of d-orbitals (t2g and eg) is small enough that it's easier for electrons to occupy the higher energy orbitals rather than pair up in the lower energy ones. In a low-spin complex, the energy gap is large, so electrons will pair up in the lower energy orbitals before occupying the higher energy ones.

Mn3+ has a d4 configuration. In a high-spin complex, all four of these electrons will be unpaired, each occupying a separate d-orbital. In a low-spin complex, two of these electrons will pair up in the lower energy orbitals, leaving only two unpaired. Thus, the difference in the number of unpaired electrons between the high-spin and low-spin complexes is indeed two.