A planet (P_1) is moving around the star of mass 2M in the orbit of...

Using Kepler's third law, the time period of revolution of a planet is given by $T = 2\pi \sqrt{\dfrac{r^3}{GM_s}}$.

For planet $P_1$, the orbital radius is $r = R$ and the mass of the star is $M_s = 2M$:
$T_1 = 2\pi \sqrt{\dfrac{R^3}{G(2M)}}$

For planet $P_2$, the orbital radius is $r = 2R$ and the mass of the star is $M_s = 4M$:
$T_2 = 2\pi \sqrt{\dfrac{(2R)^3}{G(4M)}} = 2\pi \sqrt{\dfrac{8R^3}{4GM}} = 2\pi \sqrt{\dfrac{2R^3}{GM}}$

Taking the ratio of $T_2$ and $T_1$:
$\dfrac{T_2}{T_1} = \dfrac{\sqrt{\dfrac{2R^3}{GM}}}{\sqrt{\dfrac{R^3}{2GM}}} = \sqrt{2 \times 2} = 2$

Answer: $2$