Answer
$\approx 1.25\;\rm cm$
Work Step by Step
We know that the beat frequency is given by
$$f_{\rm Beat}=|f_1-f_2|$$
A longer wavelength means less frequency. This means that $f_1\gt f_2$.
We also know that the speed of the microwave is the speed of light and it is given by $c=\lambda f$, so $f=c/\lambda$
$$f_{\rm Beat}=f_1-f_2=\dfrac{c}{\lambda_1}-\dfrac{c}{\lambda_2}$$
Solving for $\lambda_2$
$$f_{\rm Beat}-\dfrac{c}{\lambda_1}=-\dfrac{c}{\lambda_2}$$
$$\lambda_2 =\dfrac{c}{\dfrac{c}{\lambda_1}-f_{\rm Beat}}$$
Plugging the known;
$$\lambda_2 =\dfrac{(3\times 10^8)}{ \dfrac{(3\times 10^8)}{ (1.250\times 10^{-2})}-100}=\bf 0.01250000005208\;\rm m$$
$$\lambda_2\approx \color{red}{\bf 1.25}\;\rm cm$$
