Answer
a) $915\;\rm m$
b) $0.152\;\rm W/m^2$
Work Step by Step
$$\color{blue}{\bf [a]}$$
The width of the central maximum, in a circular aperture, is given by
$$w=\dfrac{2.44\lambda L}{D}$$
This width is the diameter of the radar beam at a distance of $L$.
Recalling that $c=\lambda f$, hence $\lambda=c/f$;
$$w=\dfrac{2.44c L}{fD}$$
Plugging the known;
$$w=\dfrac{2.44(3\times 10^8)(30\times 10^3)}{(12\times 10^9)(2)}$$
$$w=\color{red}{\bf 915}\;\rm m$$
$$\color{blue}{\bf [b]}$$
The average intensity is given by
$$I=\dfrac{P}{A}$$
where $A$ is the area and $P$ is the power.
Hence,
$$I=\dfrac{P}{\pi r^2}$$
Plugging the known;
$$I=\dfrac{100\times 10^3}{\pi \left(\frac{915}{2}\right)^2}$$
$$I=\color{red}{\bf 0.152}\;\rm W/m^2$$
