1. Find the oxidation state of Co in $\left[\mathrm{Co}(\mathrm{H_2O})_6\right]^{3+}$

   • Water is a neutral ligand.
   • Let oxidation state of Co be $x$.

   $x + 6(0) = +3 \Rightarrow x = +3$

   So, the metal ion is $\mathrm{Co}^{3+}$.

2. Write the electronic configuration of Co and $\mathrm{Co}^{3+}$

   • Atomic number of Co = 27
   • Ground state of Co:

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

   • For $\mathrm{Co}^{3+}$, remove 2 electrons from $4s$ and 1 electron from $3d$:

   $\mathrm{Co}^{3+}: [\mathrm{Ar}]\,3d^6$

3. Determine the nature of the ligand field

   • Ligand = $\mathrm{H_2O}$
   • For $\mathrm{Co}^{3+}$, even with $\mathrm{H_2O}$, the octahedral splitting is sufficiently large to give a low-spin complex.

   Thus, for octahedral $d^6$ low spin:

   $t_{2g}^6 e_g^0$

4. Count unpaired electrons

   In $t_{2g}^6 e_g^0$, all 6 electrons are paired.

   Therefore, number of unpaired electrons = $0$.

5. Check options

   • A: 0 ✅
   • B: 2 ❌
   • C: 1 ❌
   • D: 4 ❌

Hence, the correct answer is A.