 Solution

Propagation through an already formed cold plasma requires ${\omega }_L>{\omega }_p$, hence $\boxed{{\lambda }_L<{\lambda }_p=2\pi c/{\omega }_p}$. For the illustrative density above, that is a cutoff near 126 micrometres. It is a propagation ceiling, not a sufficient condition for producing ionization.

TODO — the unique cutoff inherits the unspecified density, and a literal maximum wavelength for gas ionization also needs gas species, laser intensity and the ionization mechanism. A single-photon threshold would impose $hc/\lambda \ge I_{\mathrm{ion}}$; strong-field or multiphoton ionization is a different criterion. None of those data is supplied. The density-dependent propagation condition is the part determined symbolically.