When I talk to non-specialists and mention that the Planck scale is where quantum gravity is likely to become relevant sometimes people get suspicious about this type of argument. If I have time, I explain that to probe smaller length details I would need so much CM energy that I create a black home and thus still cannot resolve it. However, if I have less time, I just say: Look, it's relativistic, gravity and quantum, so it's likely that c, G and h play a role. Turn those into a length scale and there is the Planck scale.
If they do not believe this gives a good estimate I ask them to guess the size of an atom: Those are quantum objects, so h is likely to appear, the binding is electromagnetic, so e (in SI units in the combination e^2/4 pi epsilon_0) has to play a role and it comes out of the dynamics of electrons, so m, the electron mass, is likely to feature. Turn this into a length and you get the Bohr radius.
Of course, as all short arguments, this has a flaw: there is a dimensionless quantity around that could spoil dimension arguments: alpha, the fine-structure constant. So you also need to say, that the atom is non-relativistic, so c is not allowed to appear.
You could similarly ask for a scale that is independant of the electric charge, and there it is: Multiply the Bohr radius by alpha and you get the electron Compton wavelength h/mc.
You could as well ask for a classical scale which should be independent of h: Just multiply another power of alpha and you get the classical electron radius e^2/4 pi epsilon_0 m c^2. At the moment, however, I cannot think of a real physical problem where this is the characteristic scale (NB alpha is roughly 1/137, so each scale is two orders of magnitude smaller than the previous).
Update: Searching Google for "classical electron radius" points to scienceworld and wikipedia, both calling it the "Compton radius". Still, there is a difference of an alpha between the Compton wavelength and the Compton radius.