= Solution
The emitted <photon energy> is the difference of the two <energy levels>:
$$
\Delta E=E_3-E_2=\frac{(9-4)\hbar^2}{2mr^2}=\frac{5\hbar^2}{2mr^2}.
$$
Using $E_\gamma=hc/\lambda$ gives
$$
\boxed{\lambda=\frac{2hcmr^2}{5\hbar^2}=\frac{4\pi cmr^2}{5\hbar}.}
$$
No numerical <mass> or <radius> is supplied, so this is the determinate result. It is the wavelength conditional on the transition occurring. An isolated <particle on a circle> with no coupling would remain in its <energy eigenstate>; a charged particle coupled to <electromagnetic radiation> can emit subject to its transition <selection rules>. The $3\to2$ change has $\Delta n=-1$, compatible with an electric-dipole angular factor for a charged ring particle.
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