Answer
a) $3.72\times10^{24}\;\rm kg$
b) $5.31\times 10^6\;\rm m$
Work Step by Step
You were on the planet Physics when you toss the rock up. So, the free-fall acceleration on this planet is given by
$$y_f=y_i+v_{iy}t+\frac{1}{2}a_yt^2$$
The initial and final height is your and, so $y_f-y_i=0$
$$0=v_{iy}t-\frac{1}{2}gt^2$$
$$g=\dfrac{2v_{iy}\color{red}{\bf\not}t}{t^{\color{red}{\bf\not}2}}=\dfrac{2v_{iy}}{t}$$
Plugging the known;
$$g =\dfrac{2(11)}{2.5}=\bf 8.8\;\rm m/s^2$$
Now we can use Newton's gravitational law exerted on the rock on its surface to find the mass of the planet.
$$\sum F_r=F_g=mg$$
$$\dfrac{GM\color{red}{\bf\not}m}{R^2}=\color{red}{\bf\not}mg$$
where $m$ is the mass of the rock and $M$ is the mass of the planet.
$$\dfrac{GM }{R^2}= g\tag 1$$
Now we are given the period of your ship around the planet Physics which is 230 minutes and we know that its distance from the center of the planet is $r=2R$.
So according to Kepler's third law,
$$T^2=\left(\dfrac{4\pi^2}{GM}\right)r^3$$
Hence,
$$T^2=\left(\dfrac{4\pi^2}{GM}\right)(2R)^3=\left(\dfrac{4\pi^2}{GM}\right)(8R^3)$$
$$T^2 = \dfrac{32\pi^2R^3}{GM} $$
Noting that $GM/R^2=g$, so
$$T^2 = \dfrac{32\pi^2R }{g } $$
So, the planet's radius is given by
$$R = \dfrac{gT^2 } {32\pi^2} $$
Plugging the known:
$$R = \dfrac{8.8(230\times 60)^2 } {32\pi^2} $$
$$R=\color{red}{\bf 5.31\times 10^6}\;\rm m\tag b$$
Solving (1) for $M$;
$$M=\dfrac{gR^2}{G}$$
Plugging the known;
$$M=\dfrac{8.8(5.31\times 10^6)^2}{6.67\times 10^{-11}}$$
$$M=\color{red}{\bf 3.72\times10^{24}}\;\rm kg\tag a$$
