Answer
$\rm \left(8.25\times 10^{10}\;\rm m, 325\;s\right)$
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.
We need to use Lorentz transformation for space.
$$x=\gamma(x'+vt')$$
where $\gamma=\left[1-\frac{v^2}{c^2}\right]^{-1/2}=\left[1-\frac{0.6^2\color{red}{\bf\not}c^2}{\color{red}{\bf\not}c^2}\right]^{-1/2}=\bf \frac{5}{4}$, and $v=0.6c$
Hence,
$$x =\frac{5}{4}(x'+0.6ct')$$
Plug the known;
$$x =\frac{5}{4}([3\times 10^{10}]+[0.6\times 3\times 10^8][200])$$
$$x=\color{red}{\bf 8.25\times 10^{10}}\;\rm m$$
Using Lorentz transformation for time.
$$t=\gamma\left[t'+\dfrac{vx'}{c^2}\right] =\frac{5}{4}\left[t'+\dfrac{0.6 cx'}{c^2}\right] $$
$$t = \frac{5}{4}\left[t'+\dfrac{0.6 x'}{c }\right] $$
Plug the known;
$$t = \frac{5}{4}\left[(200)+\dfrac{0.6 (3\times 10^{10})}{(3\times 10^8) }\right] $$
$$t=\color{red}{\bf 325}\;\rm s$$
