Step 1: Understand the clue given.
The highest third ionization enthalpy belongs to the element whose $M^{2+}$ ion has a specially stable arrangement, because removing the third electron from a stable ion is very hard.
Step 2: Test manganese.
Manganese is $[Ar]3d^5 4s^2$. After losing two electrons, $Mn^{2+}$ is $[Ar]3d^5$, a half-filled $d$ subshell that is extra stable.
Step 3: Explain the high value.
Because $Mn^{2+}$ ($3d^5$) is so stable, pulling out the third electron needs the most energy. So Mn has the highest third ionization enthalpy among the listed metals.
Step 4: Find the +3 configuration.
Removing one more electron gives $Mn^{3+} = [Ar]3d^4$. This $d^4$ ion has $4$ unpaired electrons.
Step 5: Use the spin-only formula.
The spin-only magnetic moment is $\mu = \sqrt{n(n+2)}$ with $n = 4$: \[ \mu = \sqrt{4(4+2)} = \sqrt{24}. \]
Step 6: Compute and conclude.
\[ \mu = \sqrt{24} \approx 4.90 \ \text{BM}. \] So the magnetic moment is \[ \boxed{4.90 \ \text{BM}} \]