Colours in Coordination Compounds
I have read that $\ce{NH3}$ acts as an SFL with a $\ce{M^3+}$ metal ion and acts as a WFL with a $\ce{M^2+}$ metal ion.

(I assume that SFL = strong-field ligand and WFL = weak field ligand.)
No, this is not correct. Ammonia is neither a true strong-field ligand nor a true weak-field ligand. Instead, it can form high-spin and low-spin complexes depending on the properties of the metal. Thus, it is best to term it a medium-field ligand.
It is also not true that all metal(III) ammine complexes be low-spin while all metal(II) ammine complexes be high-spin. However, among the many different factors that influence the spin state of a complex, it is true that a higher oxidation state tends to favour (note the wording) a low-spin configuration.

It also states that when $\ce{NH3}$ forms a complex with the $\ce{M^2+}$ ion, its stability constant is $10^{11}$.

It would be very weird if the stability constants of all $\ce{[M(NH3)6]^2+}$ complexes have the same value. Think about it: the metals are different while everything else is the same. There has to be variation. The same is true for the aquacomplexes.

So, if that is true then $\ce{[Ni(H2O)6]^2+}$ should be more stable than $\ce{[Ni(NH3)6]^2+}$ as Ni is in the $\mathrm{+II}$ oxidation state.

You can’t say much about the stability of the two a priori. What you can say is that both of these complexes should be relatively inert (although not as inert as, say, $\ce{[Co(NH3)6]^3+}$) from a simple CFSE estimation.

The crux of the question is that we must arrange the complexes in correct order for the absorption wavelength in the visible region

This has absolutely nothing to do with the stability of the complexes or their stability constant; only with their electronic structure and the energy difference between the orbital an electron is excited from and the orbital an electron is excited into. Furthermore, there has to be an error in $\ce{[Ni(NO2)6]^2+}$ – I find it hard to see a stable complex involving six stable radicals around it.
As for the order the question asks for, this is where the spectrochemical series actually comes in and in handy. Determine which orbitals are involved, what effect a greater field split has on these, find the ligands on the spectrochemical series, determine their relative field splits and translate that all into an photon wavelength. You will soon notice that the wavelength of the hexaaquanickel(II) complex must be longer than the wavelength of hexaamminenickel(II) which gives you your answer by inferring.