Answer
a) $\approx 192\;\rm m$
b) $2.17\times10^5\;\rm m$
Work Step by Step
We know, for diffraction, that
$$\sin\theta=\dfrac{1.22\lambda}{d}$$
where $\theta$ is the angle that represents half the angle of the laser beam spread. and for small angles $\sin\theta\approx \theta$
Thus, from the figure below,
$$\theta_{beam}=2\theta=2\times\dfrac{1.22\lambda}{d}$$
Plugging the known;
$$\theta_{beam} =2\times \dfrac{1.22\times 694\times10^{-9}}{3\times10^{-3}}$$
$$\theta_{beam}=\color{blue}{\bf5.644\times 10^{-4}}\;\rm rad$$
a) In this case, at the satellite, $r=340\;000$ m.
Thus the diameter of the beam at this distance is given by
$$D=r\theta=340\;000\times 5.644\times 10^{-4}$$
$$D\approx \color{red}{\bf192}\;\rm m$$
b) In this case, at the moon, $r=384\times10^6$ m.
Thus the diameter of the beam at this distance is given by
$$D=r\theta=384\times10^6\times 5.644\times 10^{-4}$$
$$D\approx \color{red}{\bf2.17\times 10^5}\;\rm m$$
