Answer
${\bf 5.24\times 10^{-6}}\;\rm N/m$
Work Step by Step
First, we need to find the intensity of the airport's transmission broadcast at the airplane position 30 km away.
Assuming that there is no waste of energy of the beam in the atmosphere.
$$I_{\rm airplane}=\dfrac{P}{A}=\dfrac{P}{4\pi r^2}$$
Plug the known;
$$I_{\rm airplane}=\dfrac{(150\times 10^3)}{4\pi (30\times 10^3)^2}=\bf 1.33\times 10^{-5}\;\rm W/m^2$$
Since the airplane scatters microwaves uniformly in all directions, it works now as a new source of microwaves.
The power of this new source is
$$P_{\rm new}=I_{\rm airplane}A_{\rm airplane}=(1.33\times 10^{-5})(31)$$
$$P_{\rm new}=\bf 4.11\times 10^{-4}\;\rm W$$
Now we need to find the intensity of the new source, reflected from the airplane, at the airport.
$$I_{\rm airport}=\dfrac{P_{\rm new}}{4\pi r^2}=\dfrac{4.11\times 10^{-4}}{4\pi (30\times 10^3)^2}=\bf 3.64\times 10^{-14}\;\rm W/m^2$$
Recalling that the intensity is also given by
$$I=\dfrac{c\epsilon_0E_0^2}{2}$$
So, in our case, the electric field strength of the microwave signal received back at the airport is given by
$$E_0=\sqrt{\dfrac{2I_{\rm airport}}{c\epsilon_0}}$$
Plug the known;
$$E_0=\sqrt{\dfrac{2(3.64\times 10^{-14}) }{(3\times 10^8)(8.85\times 10^{-12})}}$$
$$E_0=\color{red}{\bf 5.24\times 10^{-6}}\;\rm N/m$$
