The third ionization enthalpy refers to the energy required to remove the third electron from a di-positive ion ($M^{2+}$) to form a tri-positive ion ($M^{3+}$). In the case of transition metals, the energies involved in removing the third electron are typically higher than for the first and second electrons due to an increasing effective nuclear charge and the electrons being removed from closer energy levels to the nucleus or from a half-filled or full-filled d-subshell which is relatively more stable.

Let us look at the electronic configurations of the ions of each metal listed to determine which would require the highest third ionisation enthalpy. Since we are talking about $3^{\text{rd}}$ ionization, we would be considering the removal of three electrons, ending up with a $+3$ oxidation state for each metal.

• Manganese ($\mathrm{Mn}$): [Ar] $3d^5 4s^2$


• Iron ($\mathrm{Fe}$): [Ar] $3d^6 4s^2$


• Chromium ($\mathrm{Cr}$): [Ar] $3d^5 4s^1$


• Vanadium ($V$): [Ar] $3d^3 4s^2$

After removing two electrons (assuming they come from the 4s orbital and the next available d orbital), the configurations would be:

• $\mathrm{Mn^{2+}}$: [Ar] $3d^5$


• $\mathrm{Fe^{2+}}$: [Ar] $3d^6$


• $\mathrm{Cr^{2+}}$: [Ar] $3d^4$


• $V^{2+}$: [Ar] $3d^3$

Now we look at the energy required to remove the third electron from the $M^{2+}$ ions:

• For $\mathrm{Mn^{2+}}$, removing the third electron would break the half-filled $3d^5$ configuration and require significant energy.
• For $\mathrm{Fe^{2+}}$, removing the third electron would form half-filled $3d^5$ configuration, which forms stable configuration.
• For $\mathrm{Cr^{2+}}$, the third electron removal will disrupt the $3d^4$ configuration, which is not particularly stable.
• For $V^{2+}$, the third electron removal will disrupt the $3d^3$ configuration.

In terms of stability, the half-filled $d^5$ subshell in manganese ($\mathrm{Mn^{2+}}$) provide extra stability. So, the $3^{\text{rd}}$ ionisation enthalpy of $\mathrm{Mn^{2+}}$ is the highest because it involves breaking a half-filled configuration which is more stable.

Thus, out of the given options, manganese ($\mathrm{Mn}$) would have the highest third ionization enthalpy. Therefore, the correct answer is:

Option A : $\mathrm{Mn}$