1. Find oxidation states and electronic configurations

   • For $\left[\mathrm{Co}(\mathrm{H_2O})_6\right]\mathrm{Cl}_2$:

     • Water is a neutral ligand.
     • Two chloride ions are outside the coordination sphere, so the complex cation is $\left[\mathrm{Co}(\mathrm{H_2O})_6\right]^{2+}$.
     • Hence, oxidation state of Co is $+2$.
     • $\mathrm{Co} : [\mathrm{Ar}]3d^74s^2$, so $\mathrm{Co}^{2+} : 3d^7$.

   • For $\left[\mathrm{Cr}(\mathrm{H_2O})_6\right]\mathrm{Cl}_3$:

     • Three chloride ions are outside, so the complex cation is $\left[\mathrm{Cr}(\mathrm{H_2O})_6\right]^{3+}$.
     • Hence, oxidation state of Cr is $+3$.
     • $\mathrm{Cr} : [\mathrm{Ar}]3d^54s^1$, so $\mathrm{Cr}^{3+} : 3d^3$.

2. Determine number of unpaired electrons

   Since $\mathrm{H_2O}$ is a weak field ligand, both complexes are high-spin octahedral.

   • $\mathrm{Co}^{2+}$ is $d^7$ in octahedral field: $t_{2g}^5 e_g^2$ Number of unpaired electrons, $n = 3$.

   • $\mathrm{Cr}^{3+}$ is $d^3$ in octahedral field: $t_{2g}^3 e_g^0$ Number of unpaired electrons, $n = 3$.

3. Calculate spin-only magnetic moments

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

   • For $\left[\mathrm{Co}(\mathrm{H_2O})_6\right]^{2+}$: $\mu = \sqrt{3(3+2)} = \sqrt{15}\, \text{BM}$

   • For $\left[\mathrm{Cr}(\mathrm{H_2O})_6\right]^{3+}$: $\mu = \sqrt{3(3+2)} = \sqrt{15}\, \text{BM}$

4. Find the difference

   $\sqrt{15} - \sqrt{15} = 0$

Therefore, the required integer answer is: $\boxed{0}$