1. Identify the starting complex and metal oxidation state

The given complex is: $[Co(H_2O)_6]^{2+}$

Since water is a neutral ligand, the oxidation state of cobalt is: $x+6(0)=+2 \Rightarrow x=+2$

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

Its electronic configuration is: $\text{Co}: [Ar]3d^74s^2$ $\text{Co}^{2+}: [Ar]3d^7$

2. Effect of excess ammonia and oxygen

Ammonia first replaces water ligands, and in presence of oxygen, cobalt(II) is oxidized to cobalt(III). The well-known product formed is an ammine complex of cobalt(III), such as: $[Co(NH_3)_6]^{3+}$

3. Why is the product diamagnetic?

For cobalt(III): $\text{Co}^{3+} = 3d^6$

In an octahedral field with $NH_3$ ligands, cobalt(III) forms a low-spin complex because $\text{Co}^{3+}$ has high crystal field splitting.

Thus the electron distribution is: $t_{2g}^6 e_g^0$

All electrons are paired, so the complex is diamagnetic, matching the statement in the question.

4. Count electrons in $t_{2g}$ orbitals

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

the number of electrons in the $t_{2g}$ orbitals is: $6$

Therefore, the required integer is: $\boxed{6}$