The correct order of spin-only magnetic moment for the given complexes is:

The spin-only magnetic moment ($\mu$) is given by the formula $\mu = \sqrt{n(n+2)}$ BM, where $n$ is the number of unpaired electrons. A higher number of unpaired electrons results in a higher magnetic moment.

1. $[MnCl_6]^{3-}$: The oxidation state of Mn is +3. The electronic configuration of $Mn^{3+}$ is $3d^4$. Since $Cl^-$ is a weak field ligand, the crystal field splitting energy is small ($\Delta_o < P$), leading to a high-spin complex ($t_{2g}^3 e_g^1$). The number of unpaired electrons is $n = 4$.

2. $[Co(H_2O)_6]^{2+}$: The oxidation state of Co is +2. The electronic configuration of $Co^{2+}$ is $3d^7$. Since $H_2O$ is a weak field ligand, it forms a high-spin complex ($t_{2g}^5 e_g^2$). The number of unpaired electrons is $n = 3$.

3. $[Fe(CN)_6]^{3-}$: The oxidation state of Fe is +3. The electronic configuration of $Fe^{3+}$ is $3d^5$. Since $CN^-$ is a strong field ligand, the crystal field splitting energy is large ($\Delta_o > P$), leading to a low-spin complex ($t_{2g}^5 e_g^0$). The number of unpaired electrons is $n = 1$.

The order of the number of unpaired electrons is $4 > 3 > 1$. Therefore, the correct order for the spin-only magnetic moment is $[MnCl_6]^{3-} > [Co(H_2O)_6]^{2+} > [Fe(CN)_6]^{3-}$.