1. Identify the coordination geometry

The complex is $[Co(ox)_2(Br)(NH_3)]^{2-}$ where each $ox$ (oxalate) is a bidentate ligand.

• Two oxalate ligands occupy $2\times 2 = 4$ coordination sites.
• $Br^-$ and $NH_3$ are monodentate ligands, occupying 1 site each.

So total coordination number is 6, hence the complex is octahedral.

2. Type of octahedral complex

This is of the form: $[M(AA)_2ab]$ where:

• $AA$ = symmetrical bidentate ligand (oxalate)
• $a = Br$
• $b = NH_3$

For octahedral complexes of type $[M(AA)_2ab]$, geometrical isomerism is possible depending on whether $a$ and $b$ are cis or trans.

3. Geometrical isomers

(i) cis isomer

Here $Br$ and $NH_3$ are adjacent.

For $[M(AA)_2ab]$, the cis form is optically active because it lacks a plane/center of symmetry. Hence it exists as a pair of enantiomers:

• one $\Delta$ form
• one $\Lambda$ form

So cis gives 2 stereoisomers.

(ii) trans isomer

Here $Br$ and $NH_3$ are opposite to each other.

The trans form has sufficient symmetry and is optically inactive, so it gives only 1 stereoisomer.

4. Total number of stereoisomers

Therefore, $\text{Total stereoisomers} = 2 \text{ (cis enantiomers)} + 1 \text{ (trans)} = 3$

5. Final answer

$\boxed{3}$

6. Comparison with stored correct answer

Stored correct answer = $3$

Our derived answer = $3$

So the stored answer is correct.