The second ionisation enthalpy is highest for Mn, because the second ionisation removes an electron from a stable [Ar]3$d^5$ configuration, making it highly stable. The spin-only magnetic moment is calculated using the formula:
\(\mu = \sqrt{n(n+2)}{ BM} \)
For Mn, with $n = 5$ (because 3$d^5$ has 5 unpaired electrons), the magnetic moment is:
\( \mu = \sqrt{5(5+2)} = \sqrt{35} \approx 6 { BM}.\)
Thus, the correct answer is (6).