Answer
See the detailed answer below.
Work Step by Step
Let's assume that at $t=0$, the two frames, $S$ for Earth and $S'$ for the rocket, are overlapping at the origin point (0,0) of the $S$ frame which is for the observer on Earth. In other words, the two origins of the two frames are at the same point at $t=0$ s.
$$\color{blue}{\bf [a]}$$
$\Rightarrow$ For the tree at the origin $x=0$ m;
We need to use Lorentz transformation for space.
$$x'=\gamma(x-vt)$$
where $\gamma=\left[1-\frac{v^2}{c^2}\right]^{-1/2}=\left[1-\frac{0.5^2\color{red}{\bf\not}c^2}{\color{red}{\bf\not}c^2}\right]^{-1/2}=\bf \frac{2\sqrt3}{3}$, and $v=0.5c$
Hence,
$$x'=\frac{2\sqrt3}{3}(x-0.5ct)$$
Plug the known;
$$x'=\frac{2\sqrt3}{3}([0]-[0.5\times 3\times 10^8][10\times 10^{-6}])$$
$$x'=\color{red}{\bf -1732}\;\rm m$$
Using Lorentz transformation for time.
$$t'=\gamma\left[t-\dfrac{vx}{c^2}\right] =\frac{2\sqrt3}{3}\left[t-\dfrac{0.5 cx}{c^2}\right] $$
$$t'= \frac{2\sqrt3}{3}\left[t-\dfrac{0.5 x}{c }\right] $$
Plug the known;
$$t'= \frac{2\sqrt3}{3}\left[(10\times 10^{-6})-\dfrac{0.5 (0)}{(3\times 10^8) }\right] $$
$$t'=\color{red}{\bf 11.55}\;\rm \mu s$$
$\Rightarrow$ For the pole at $x=30\times 10^3$ m;
We need to use Lorentz transformation for space.
$$x'=\gamma(x-vt)$$
where $\gamma=\left[1-\frac{v^2}{c^2}\right]^{-1/2}=\left[1-\frac{0.5^2\color{red}{\bf\not}c^2}{\color{red}{\bf\not}c^2}\right]^{-1/2}=\bf \frac{2\sqrt3}{3}$, and $v=0.5c$
Hence,
$$x'=\frac{2\sqrt3}{3}(x-0.5ct)$$
Plug the known;
$$x'=\frac{2\sqrt3}{3}([30\times 10^3]-[0.5\times 3\times 10^8][10\times 10^{-6}])$$
$$x'=\color{red}{\bf 32.91}\;\rm km$$
Using Lorentz transformation for time.
$$t'=\gamma\left[t-\dfrac{vx}{c^2}\right] =\frac{2\sqrt3}{3}\left[t-\dfrac{0.5 cx}{c^2}\right] $$
$$t'= \frac{2\sqrt3}{3}\left[t-\dfrac{0.5 x}{c }\right] $$
Plug the known;
$$t'= \frac{2\sqrt3}{3}\left[(10\times 10^{-6})-\dfrac{0.5 (30\times 10^3)}{(3\times 10^8) }\right] $$
$$t'=\color{red}{\bf -46.19}\;\rm \mu s$$
$$\color{blue}{\bf [b]}$$
According to the results above, it is obvious that the two events are not instantaneous in the rocket frame of reference since the strike on the pole seems to have occurred before the tree strike by a time of
$$t=46.19+11.55=\bf 57.74\;\rm \mu s$$
