To solve this problem, we need to understand what factors influence the second ionisation enthalpy of transition metals and how to calculate the spin-only magnetic moment of their ions. The second ionization enthalpy refers to the energy required to remove an electron from the singly charged cation of an element.

Generally, for transition metals, the removal of an electron from a d-orbital requires less energy compared to removing an electron from an s-orbital of a valence shell due to the penetration effect and the shield. However, among the given options, the highest second ionisation enthalpy would correspond to the element whose cation (after losing one electron) attains a half-filled or completely filled d-orbital configuration due to the extra stability associated with such electronic arrangements.

Let's examine the electronic configurations of the cations formed after the first ionization (this helps in understanding the stability and hence the energy required for the second ionization):

Sc ($$Z = 21$$) loses one electron: [Ar] $$3d^1$$
Ti ($$Z = 22$$) loses one electron: [Ar] $$3d^2$$
V ($$Z = 23$$) loses one electron: [Ar] $$3d^3$$
Cr ($$Z = 24$$) has an anomalous configuration of [Ar] $$3d^5$$ 4s^1 in its neutral atom, loses one electron to become [Ar] $$3d^5$$
Mn ($$Z = 25$$) loses one electron: [Ar] $$3d^5$$
Fe ($$Z = 26$$) loses one electron: [Ar] $$3d^6$$

Among these configurations, $$\mathrm{Mn}^+$$ ([Ar] $$3d^5$$) has a half-filled d-orbital, lending it extra stability due to exchange energy and symmetry. Consequently, $$\mathrm{Mn}$$ has the highest second ionization enthalpy among the given elements, as it would require more energy to remove an electron from a stable $$\mathrm{Mn}^+$$ ion.

Now, to find the spin-only magnetic moment of $$\mathrm{Mn}^+$$, we use the formula:

$$\mu = \sqrt{n(n + 2)}\, \text{B.M.}$$

where $$n$$ is the number of unpaired electrons. In the case of $$\mathrm{Mn}^+$$, the electronic configuration is [Ar] $$3d^5$$, which means all five d-electrons are unpaired.

Thus, $$n = 5$$, and substituting in the formula for magnetic moment:

$$\mu = \sqrt{5(5 + 2)} = \sqrt{35} \approx 5.92\, \text{B.M.}$$

Therefore, the spin-only magnetic moment value of $$\mathrm{Mn}^+$$ ion is approximately 5.92 BM, and the nearest integer would be 6 BM.