Using Kepler's third law, the time period of revolution of a planet is given by T=2πGMsr3.
For planet P1, the orbital radius is r=R and the mass of the star is Ms=2M:
T1=2πG(2M)R3
For planet P2, the orbital radius is r=2R and the mass of the star is Ms=4M:
T2=2πG(4M)(2R)3=2π4GM8R3=2πGM2R3
Taking the ratio of T2 and T1:
T1T2=2GMR3GM2R3=2×2=2
Answer: 2