I don't understand what this radius means. Aren't orbits elliptical, or are orbits about a center of mass actually circular?
I think that you will have to assume a circular orbit - otherwise, "radius" has no meaning as you correctly point out.
As for the main body of your question - I think you are making this harder than it needs to be.
You are asked to compute the radius in terms of Earth's radius. Now we know from Kepler's Laws that
$$T^2 GM = 4 \pi^2 r^3$$
According to this article on gravitational two-body problems, for the case where the mass of the planet is not negligible compared to the mass of the "sun", we simply replace $M$ with $M+m$; now we can write down the first equation for the mass and orbital radius of the planet:
$$\frac{T_e^2 G M_s}{r_e^3} = \frac{T_p^2~ G ~(0.9~ M_s + m_p)}{r_p^3}\tag1$$
There are only two unknowns in this equation: $r_p$ and $m_p$.
Next, since you have the line of sight velocity and the period, the distance $r_p$ of the planet to the center of rotation follows immediately:
$$\begin{align}v &= \omega r_p = \frac{2\pi r_p}{T}\\
\Rightarrow r_p &= \frac{vT}{2\pi}\end{align}\tag2$$
Now that you have $r_p$ you can substitute into (1) and that should give you your answer.
Converting these to the radius of Earth and the mass of Jupiter should be straightforward. It might be worth calculating how different the answer would be if you could assume the planet to be "light" - then the simple case of Kepler's law would tell us
$$\frac{r_p}{r_e}=\left(\frac{1500}{365.25}\right)^{\frac23}\approx 2.56$$
This is not the answer to your question - but how far off is it from the answer you get from the above? Before you do the calculation askyourself this - do you expect the number to be bigger or smaller?
$R$ is the distance between the centres of either object. By centres I mean their centres of mass (the point their gravitational forces "pull" from, so to speak).
$R$ is thus the sum of Earth's radius $R_e$, the object's radius $r_o$ (if spherical) and the height $h$ it is located above the ground:
$$R=R_e+h+r_o$$
For small objects not-too-far from the ground (such as thrown stones, fired rockets and orbiting satellites) you'll often see the radius of the object and the distance neglected. A rock's or a satellite's small radius is negligible compared to the ~6400 km of the Earth. Adding maybe 1km or even 10km or 100km in distance to this number makes no practical difference.
$$R=R_e+h+r_o\approx R_e$$
But when calculating escape velocity of e.g. Pluto's moon from Pluto, two objects that are comparable in size, you can't neglect either radius. And when calculating escape velocity of our own Moon from Earth, the ~400 000km distance must certainly be included as well (the sizes are almost negligible compared to this).
Best Answer
The easy way to calculate the escape velocity is to use the fact that for an object to just escape a gravitational well its total energy must be zero. That is, the negative gravitational potential energy and the positive kinetic energy sum to zero. So if we take the Earth as an example we have the gravitational potential energy:
$$ V = -\frac{GM_em}{r_e} $$
and the kinetic energy:
$$ T = \tfrac{1}{2}mv^2 $$
and requiring these sum to zero gives:
$$ T + V = \tfrac{1}{2}mv^2 -\frac{GM_em}{r_e} = 0 $$
and we get the usual equation for the escape velocity:
$$ v = \sqrt{\frac{2GM_e}{r_e}} $$
The point of all this is that potential energies are simply additive. For an object on the Earth's surface we add together the potential energy due to the Sun's gravity and the potential energy due to the Earth's gravity to give a total potential energy:
$$ V_\text{tot} = -\frac{GM_\text{Sun}m}{r_\text{Sun}} - \frac{GM_\text{Earth}m}{r_\text{Earth}} $$
Simply require that the kinetic energy of the object be equal to the total potential and you have calculated the escape velocity from the Earth's surface to out of the Solar System.
You are correct to note that the Earth is already moving due to it's orbital velocity. The escape velocity relative to the Earth will of course depend on which direction you launch in. For high accuracy don't forget that the surface of the Earth is rotating, so include that velocity as well.