1. Find the oxidation state of Co in $[Co(CN)_6]^{4-}$.

   Let the oxidation state of Co be $x$.

   Each $CN^-$ ligand has charge $-1$, so: $x + 6(-1) = -4$ $x - 6 = -4$ $x = +2$

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

2. Write the electronic configuration of $Co^{2+}$.

   Atomic number of Co = 27.

   Neutral Co: $Co: [Ar] \, 3d^7 4s^2$

   For $Co^{2+}$, remove two electrons from $4s$ first: $Co^{2+}: [Ar] \, 3d^7$

3. Determine the nature of the ligand.

   $CN^-$ is a strong field ligand. Hence, in an octahedral complex it causes pairing of electrons.

4. Distribute electrons in octahedral crystal field.

   For a strong field octahedral $d^7$ configuration: $t_{2g}^6 e_g^1$

   Number of unpaired electrons, $n = 1$.

5. Calculate the spin-only magnetic moment.

   Formula: $\mu = \sqrt{n(n+2)} \, \text{BM}$

   Putting $n=1$: $\mu = \sqrt{1(1+2)} = \sqrt{3} \approx 1.73 \, \text{BM}$

6. Integer answer.

   Since the question asks for an integer value, the magnetic moment is approximately: $1.73 \approx 2 \text{ BM}$

Final Answer: $2$