1. Identify the metal with highest E^(M^{3+}/M^{2+}) among Mn, Cr, Fe, Co

   The standard reduction potentials are approximately:

   $$\mathrm{Mn}^{3+} + e^- \to \mathrm{Mn}^{2+} \quad E^\circ \approx +1.5\,\text{V}$$ $$\mathrm{Co}^{3+} + e^- \to \mathrm{Co}^{2+} \quad E^\circ \approx +1.82\,\text{V}$$ $$\mathrm{Fe}^{3+} + e^- \to \mathrm{Fe}^{2+} \quad E^\circ \approx +0.77\,\text{V}$$ $$\mathrm{Cr}^{3+} + e^- \to \mathrm{Cr}^{2+} \quad E^\circ \approx -0.41\,\text{V}$$

   Hence, the highest value is for Co.

2. Write the complex

   Therefore the complex is:

   $$[\mathrm{Co}(\mathrm{CN})_6]^{4-}$$

3. Find oxidation state of Co

   Let oxidation state of Co be $x$.

   Since each $\mathrm{CN}^-$ has charge $-1$:

   $$x + 6(-1) = -4$$ $$x - 6 = -4$$ $$x = +2$$

   So the metal ion is $\mathrm{Co}^{2+}$.

4. Find d-electron configuration of $\mathrm{Co}^{2+}$

   Atomic number of Co = 27.

   Neutral Co:

   $$[\mathrm{Ar}]\,3d^7 4s^2$$

   For $\mathrm{Co}^{2+}$, remove two electrons from $4s$:

   $$[\mathrm{Ar}]\,3d^7$$

   So it is a $d^7$ system.

5. Use ligand strength of $\mathrm{CN}^-$

   $\mathrm{CN}^-$ is a strong-field ligand, so the octahedral complex is low spin.

   Octahedral splitting:

   $$t_{2g} \text{ lower}, \quad e_g \text{ higher}$$

   For low-spin $d^7$:

   $$t_{2g}^6 e_g^1$$

6. Count electrons in $e_g$

   From

   $$t_{2g}^6 e_g^1$$

   the number of electrons in $e_g$ is:

   $$1$$

Final Answer: $\boxed{1}$