1. Find oxidation state and $d$-electron count for each complex

We test whether each octahedral complex is diamagnetic (all electrons paired).

2. Complex A: $\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-}$

Let oxidation state of Mn be $x$.

$x+6(-1)=-3$ $x=+3$

So Mn is $\mathrm{Mn}^{3+}$.

Mn: $Z=25$, ground state $[\mathrm{Ar}]3d^54s^2$

$\mathrm{Mn}^{3+} : 3d^4$

Since $\mathrm{CN}^-$ is a strong-field ligand, the octahedral complex is low spin:

$d^4 \to t_{2g}^4 e_g^0$

This still has 2 unpaired electrons, so it is paramagnetic, not diamagnetic.

So, A is not diamagnetic.

3. Complex B: $\left[\mathrm{Co}(\mathrm{NH}_3)_6\right]^{3+}$

Let oxidation state of Co be $x$.

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

So Co is $\mathrm{Co}^{3+}$.

Co: $Z=27$, ground state $[\mathrm{Ar}]3d^74s^2$

$\mathrm{Co}^{3+} : 3d^6$

In octahedral field with $\mathrm{NH}_3$ (and especially for $\mathrm{Co}^{3+}$), the complex is low spin:

$d^6 \to t_{2g}^6 e_g^0$

All electrons are paired.

So, B is diamagnetic.

4. Complex C: $\left[\mathrm{Fe}(\mathrm{CN})_6\right]^{4-}$

Let oxidation state of Fe be $x$.

$x+6(-1)=-4$ $x=+2$

So Fe is $\mathrm{Fe}^{2+}$.

Fe: $Z=26$, ground state $[\mathrm{Ar}]3d^64s^2$

$\mathrm{Fe}^{2+} : 3d^6$

$\mathrm{CN}^-$ is a strong-field ligand, so octahedral complex is low spin:

$d^6 \to t_{2g}^6 e_g^0$

All electrons are paired.

So, C is diamagnetic.

5. Complex D: $\left[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_3\mathrm{F}_3\right]$

Here $\mathrm{H_2O}$ is neutral and $\mathrm{F}^-$ is $-1$ each. Since the complex is neutral:

$x+3(0)+3(-1)=0$ $x=+3$

So Co is $\mathrm{Co}^{3+}$.

$\mathrm{Co}^{3+} : 3d^6$

But $\mathrm{H_2O}$ and $\mathrm{F}^-$ are weak-field ligands, so this is high spin octahedral:

$d^6 \to t_{2g}^4 e_g^2$

This has 4 unpaired electrons, so it is paramagnetic.

So, D is not diamagnetic.

6. Conclusion

Only B and C are diamagnetic.

Therefore, the correct option is:

$\boxed{\text{A: B and C Only}}$

7. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

They agree.