The third ionization enthalpy is minimum for:

🧠 $IE_3$ is minimum for Fe because removing the third electron lands you on the stable half-filled $3d^5$ state $IE_3$ means going from $\ce{M^{2+}}$ to $\ce{M^{3+}}$. Whichever transition gives you a particularly stable product configuration will be the easiest — and a half-filled $3d^5$ has extra stability from exchange energy.

🗺️ Look at $\ce{M^{2+}}$ configurations and what they become as $\ce{M^{3+}}$ Mn ($Z=25$): $\ce{Mn^{2+}}$ is $3d^5$ (already half-filled, very stable). Removing one electron destroys this stable state — $IE_3$ is HIGH. Fe ($Z=26$): $\ce{Fe^{2+}}$ is $3d^6$. Removing one electron gives $3d^5$ — perfectly half-filled. The product is stable, so $IE_3$ is LOW. Co ($Z=27$): $\ce{Co^{2+}}$ is $3d^7$. Going to $3d^6$ — no special stability gained. Ni ($Z=28$): $\ce{Ni^{2+}}$ is $3d^8$. Going to $3d^7$ — no special stability.

So Fe uniquely benefits from landing on $3d^5$ during the third ionization. Minimum $IE_3$ is for Fe.

⚠️ The trap Many students pick Mn thinking "Mn²⁺ is $3d^5$, that's stable, so easy to ionize". This reasoning is exactly backwards. The starting state being stable means it RESISTS losing electrons (high $IE_3$). What helps is when the FINAL state ($\ce{M^{3+}}$) is the stable half-filled config. That happens for Fe, not Mn.

$\boxed{\text{Answer: (b)}}$