1. Identify metal oxidation state and d-electron count

We need an octahedral, diamagnetic, low-spin complex.

Option A: $[\text{Co}(\text{NH}_3)_6]^{3+}$

• $\text{NH}_3$ is a neutral ligand.
• Oxidation state of Co = $+3$.
• Co: $Z=27$, so $\text{Co}^{3+} = 3d^6$
• Geometry: six ligands $\Rightarrow$ octahedral.
• $\text{NH}_3$ is a relatively strong-field ligand, so for $\text{Co}^{3+}$ it gives low spin.
• Octahedral low-spin $d^6$ configuration: $t_{2g}^6 e_g^0$
• All electrons are paired $\Rightarrow$ diamagnetic.

So, A satisfies all conditions.

Option B: $[\text{NiCl}_4]^{2-}$

• Let oxidation state of Ni be $x$: $x + 4(-1) = -2 \Rightarrow x=+2$
• $\text{Ni}^{2+} = 3d^8$
• With 4 ligands and weak-field $\text{Cl}^-$, this complex is generally tetrahedral, not octahedral.
• Tetrahedral $d^8$ is paramagnetic.

So, B is not octahedral and not diamagnetic low spin.

Option C: $[\text{CoCl}_6]^{3-}$

• Oxidation state of Co: $x + 6(-1) = -3 \Rightarrow x=+3$
• $\text{Co}^{3+} = 3d^6$
• Six ligands $\Rightarrow$ octahedral.
• But $\text{Cl}^-$ is a weak-field ligand, so this is high spin.
• High-spin octahedral $d^6$: $t_{2g}^4 e_g^2$
• It has 4 unpaired electrons $\Rightarrow$ paramagnetic.

So, C is not diamagnetic low spin.

Option D: $[\text{CoF}_6]^{3-}$

• Oxidation state of Co: $x + 6(-1) = -3 \Rightarrow x=+3$
• $\text{Co}^{3+} = 3d^6$
• Six ligands $\Rightarrow$ octahedral.
• $\text{F}^-$ is a weak-field ligand, so this is high spin.
• High-spin octahedral $d^6$ is paramagnetic.

So, D is not diamagnetic low spin.

2. Conclusion

Only option A is an octahedral diamagnetic low-spin complex: $[\text{Co}(\text{NH}_3)_6]^{3+}$