__Concept__:

The cut-off frequency for a rectangular waveguide is given by:

\({f_{m,\;n}} = \frac{C}{2}\sqrt {{{\left( {\frac{m}{a}} \right)}^2} + {{\left( {\frac{n}{b}} \right)}^2}} \)

C = velocity of light in free space.

m, n = mode of operation of the waveguide

a, b = dimensions of the waveguide

__Calculation__:

For TE_{2, 1} mode, m = 2 and n = 1

Also, a = 5 cm

b = 3 cm

The cut-off frequency will be:

\({f_{c\left( {2,\;1} \right)}} = \frac{{3\; \times \;{{10}^{10}}}}{2}\sqrt {{{\left( {\frac{2}{5}} \right)}^2} + {{\left( {\frac{1}{3}} \right)}^2}} \)

\( = 1.5 \times {10^{10}}\sqrt {0.16 + 0.11} Hz\)

= 1.5 × 10^{10} × 0.52

f_{c(2, 1)} = 7810.24 MHz