Problem 7.8.3: A TM_{010} circular cavity

and $a/d=1.148$:

With $c/2\pi =4.775\mathrm{GHz}\mathrm{cm}$ and the radius in cm:

The next two are TM${}_{110}$ at $7.97\mathrm{GHz}$ and TE${}_{111}$ at $8.42\mathrm{GHz}$.

TE${}_{111}$ falls below TM${}_{010}$ when $(1.841/a{)}^2+(\pi /d{)}^2<(2.405/a{)}^2$, that is for

For longer cavities TE${}_{111}$ drops below $5\mathrm{GHz}$.

On the mode chart ($v=30\mathrm{GHz}\mathrm{cm}$) the cavity sits at $(2a/d{)}^2=5.27$, and the TM${}_{010}$ line is at $(2af{)}^2=(4.593\times 5{)}^2=527.4(\mathrm{GHz}\mathrm{cm}{)}^2$, the value $(30\times 2.405/\pi {)}^2$ of the chart. TE${}_{111}$ there: $y=309.1+225\times 5.27=1495$ and $f=\sqrt{1495}/4.593=8.42\mathrm{GHz}$, as found.