Pair of transition metal ions having the same number of unpaired electrons is:

First list the atomic numbers and ground-state electronic configurations of the elements involved (outer part only).

Ti (22): $$3d^2\,4s^2$$    V (23): $$3d^3\,4s^2$$    Cr (24): $$3d^5\,4s^1$$    Mn (25): $$3d^5\,4s^2$$    Fe (26): $$3d^6\,4s^2$$    Co (27): $$3d^7\,4s^2$$

When a transition metal forms a positive ion, electrons are removed first from the $$4s$$ orbital and then from $$3d$$.

Case 1: $$V^{2+}$$ and $$Co^{2+}$$

$$V^{2+}:$$ remove two $$4s$$ electrons ⇒ $$3d^3$$ Number of unpaired $$d$$ electrons = 3 (one in each of the first three $$d$$ orbitals).

$$Co^{2+}:$$ remove two $$4s$$ electrons ⇒ $$3d^7$$ High-spin $$d^7$$ distribution: $$t_{2g}^{5}\,e_g^{2}$$ ⇒ 3 unpaired electrons.

Both ions have 3 unpaired electrons.

Case 2: $$Ti^{2+}$$ and $$Co^{2+}$$

$$Ti^{2+}:$$ $$3d^2$$ ⇒ 2 unpaired electrons.
$$Co^{2+}:$$ 3 unpaired electrons. ⇒ Not equal.

Case 3: $$Fe^{3+}$$ and $$Cr^{2+}$$

$$Fe^{3+}:$$ remove two $$4s$$ and one $$3d$$ ⇒ $$3d^5$$ (high spin) ⇒ 5 unpaired electrons.
$$Cr^{2+}:$$ remove two $$4s$$ ⇒ $$3d^4$$ ⇒ 4 unpaired electrons. ⇒ Not equal.

Case 4: $$Ti^{3+}$$ and $$Mn^{2+}$$

$$Ti^{3+}:$$ $$3d^1$$ ⇒ 1 unpaired electron.
$$Mn^{2+}:$$ $$3d^5$$ (high spin) ⇒ 5 unpaired electrons. ⇒ Not equal.

The only pair with the same number of unpaired electrons is $$V^{2+}$$ and $$Co^{2+}$$.

Hence, the correct option is Option A.