Identify the oxidation state and d-electron count for each complex:

For $$\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+}$$:

Ethylenediamine (en) is neutral.

Oxidation state of cobalt is +3.

Cobalt (atomic number 27) in the +3 state gives a $$d^6$$ configuration.

With strong/moderate field ligands like en, the complex is low spin, leading to a configuration of $$t_{2g}^6 e_g^0$$.

For $$\left[\mathrm{CoF}_6\right]^{3-}$$:

Each fluoride ion is $$\mathrm{F}^-$$ so, with six of them, Co must be in the +3 state again.

However, since fluoride is a weak field ligand, this complex adopts a high spin configuration for $$d^6$$, resulting in $$t_{2g}^4 e_g^2$$.

For $$\left[\mathrm{Mn}\left(\mathrm{H}_2\mathrm{O}\right)_6\right]^{2+}$$:

Water is neutral.

Manganese in the +2 state gives a $$d^5$$ configuration.

With weak field water, the $$d^5$$ configuration is high spin: $$t_{2g}^3 e_g^2$$.

For $$\left[\mathrm{Zn}\left(\mathrm{H}_2\mathrm{O}\right)_6\right]^{2+}$$:

Zinc in the +2 state has a $$d^{10}$$ configuration.

In an octahedral field, all 10 electrons fill the orbitals as $$t_{2g}^6 e_g^4$$.

Calculate the crystal field stabilization energy (CFSE) for each case:

The CFSE in an octahedral field can be estimated as:

$$\mathrm{CFSE} = (n_{t_{2g}}\times -0.4\Delta_o) + (n_{e_g}\times 0.6\Delta_o)$$

For $$t_{2g}^6 e_g^0$$ (low-spin $$d^6$$ in $$\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+}$$):

$$\mathrm{CFSE} = 6(-0.4\Delta_o) + 0(0.6\Delta_o) = -2.4\Delta_o$$

For $$t_{2g}^4 e_g^2$$ (high-spin $$d^6$$ in $$\left[\mathrm{CoF}_6\right]^{3-}$$):

$$\mathrm{CFSE} = 4(-0.4\Delta_o) + 2(0.6\Delta_o) = -1.6\Delta_o + 1.2\Delta_o = -0.4\Delta_o$$

For $$t_{2g}^3 e_g^2$$ (high-spin $$d^5$$ in $$\left[\mathrm{Mn}\left(\mathrm{H}_2\mathrm{O}\right)_6\right]^{2+}$$):

$$\mathrm{CFSE} = 3(-0.4\Delta_o) + 2(0.6\Delta_o) = -1.2\Delta_o + 1.2\Delta_o = 0$$

For $$t_{2g}^6 e_g^4$$ (for $$d^{10}$$ in $$\left[\mathrm{Zn}\left(\mathrm{H}_2\mathrm{O}\right)_6\right]^{2+}$$):

$$\mathrm{CFSE} = 6(-0.4\Delta_o) + 4(0.6\Delta_o) = -2.4\Delta_o + 2.4\Delta_o = 0$$

Compare the CFSE values:

$$t_{2g}^6 e_g^0$$ gives a CFSE of $$-2.4\Delta_o$$ (most negative, hence highest stabilization).

The others result in less stabilization (or net zero).

Conclusion:

The complex with the highest CFSE is the one with the configuration $$t_{2g}^6 e_g^0$$, which corresponds to $$\left[\mathrm{Co}(\mathrm{en})_3\right]^{3+}$$.

Thus, the answer is:

Option A: $$t_{2g}^6 e_g^0$$.