= Solution
Assume a hollow, perfectly conducting <rectangular electromagnetic waveguide> filled with air approximated by vacuum. Let its wider side be $a=0.040\,\mathrm m$ and its narrower side $b=0.020\,\mathrm m$. The longitudinal propagation constant of a mode is
$$
k_z^2=\frac{\omega^2}{c^2}-\left(\frac{m\pi}{a}\right)^2-\left(\frac{n\pi}{b}\right)^2,
$$
so the <waveguide cutoff frequency> is
$$
f_{c,mn}=\frac c2\sqrt{\left(\frac ma\right)^2+\left(\frac nb\right)^2}.
$$
For a mode with <transverse electric polarization>, one of $m,n$ may vanish, but they cannot both vanish. For a mode with <transverse magnetic polarization>, both must be positive. A single hollow conducting guide has no <transverse electromagnetic mode> with zero cutoff.
The lowest cutoff therefore belongs to the $\mathrm{TE}_{10}$ mode, varying across the wider side:
$$
\boxed{f_c=\frac{c}{2a}=\frac{2.998\times10^8\,\mathrm{m\,s^{-1}}}{0.080\,\mathrm m}=3.75\times10^9\,\mathrm{Hz}\simeq3.75\,\mathrm{GHz}.}
$$
Strictly, a traveling mode transporting <energy> down the guide requires $f>f_c$. At exactly $f_c$, $k_z=0$ and the axial <group velocity> vanishes. Below cutoff the field is evanescent. A different <dielectric> filling would replace $c$ by the <wave speed> in that material.
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