R is indeed orbital Radius because:
1) Orbital radius are from the centre of masses.
2) Notice that the magnitude of order is over 10E3, such that including the mass' radius would not change the answer significantly (eg. 1000 - 1 = 999 which has less than 1% error).
Now you know F = G(M1)(M2)/R^2 and F = Ma. Now sub in and reduce and you get a = G(M)/R^2. Sub in and you get A, C, and D for answers.
As a check, you know that F is the same between the two masses, so F is constant. So using Jupiter first, we get F = (Mj)(a). Now for the sun, Ms > Mj. For F to be the same, acceleration has to decrease w.r.t Jupiter's acceleration. Is that true with the answer above? Yes.
Be glad that you're only dealing with 1D accelerations and not 3D.
