= Solution
Assume a hollow square <electromagnetic waveguide> with perfectly conducting walls, filled with <vacuum> or air. For side $a$, the <waveguide cutoff frequency> is
$$
f_{c,mn}=\frac{c}{2}\sqrt{\left(\frac m a\right)^2+\left(\frac n a\right)^2}.
$$
A transverse electric mode can have one zero index, but not both. A transverse magnetic mode requires both indices positive. Thus the lowest cutoff belongs to the two degenerate transverse electric modes $\mathrm{TE}_{10}$ and $\mathrm{TE}_{01}$, with
$$
\boxed{f_c=\frac{c}{2a}=\frac{2.9979\times10^8}{2\times10^{-2}}\,\mathrm{Hz}
\simeq15.0\,\mathrm{GHz}.}
$$
The corresponding cutoff <vacuum> <wavelength> is $2a=2.0\,\mathrm{cm}$. Strictly, a travelling mode carrying <power> along the guide needs $f>f_c$: at the cutoff itself, the axial <wavenumber> and <group velocity> are zero. The number above is therefore the lower <frequency> threshold, not a propagating mode with nonzero longitudinal <energy flux> exactly at that <frequency>. A <dielectric> filling would replace $c$ by the wave <speed> in that medium.
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