Answer
$1600\;\rm Earth\;days $
Work Step by Step
We are told that the orbit radius of planet Y around the star Omega is 4 times that of X.
$$R_Y=4R_X\tag 1$$
We know, from Kepler's laws, that the periodic time of a circular orbit is given by
$$T^2=\left(\dfrac{4\pi^2 }{GM}\right) R^3$$
where $M$ is the mass of the star.
Hence,
$$T_X^2=\left(\dfrac{4\pi^2 }{GM_{O}}\right) R_X^3$$
and
$$T_Y^2=\left(\dfrac{4\pi^2 }{GM_{O}}\right) R_Y^3$$
Finding $T_Y^2/T_X^2$
$$\dfrac{T_Y^2}{T_X^2}=\dfrac{\left(\dfrac{4\pi^2 }{GM_{O}}\right) R_Y^3}{\left(\dfrac{4\pi^2 }{GM_{O}}\right) R_X^3}=\dfrac{R_Y^3}{R_X^3}$$
Hence,
$$ T_Y^2 = \dfrac{R_Y^3T_X^2}{R_X^3}$$
Plugging from (1);
$$ T_Y^2 = \dfrac{(4R_X)^3 T_X^2}{R_X^3}= \dfrac{64 \color{red}{\bf\not}R_X^3T_X^2}{ \color{red}{\bf\not}R_X^3}$$
$$ T_Y =\sqrt{64T_X^2} =8T_X $$
Plugging the known;
$$ T_Y =8(200\;\rm Earth\;days)$$
$$T_Y=\color{red}{\bf 1600}\;\rm Earth\;days $$
Therefore, the year of planet Y is 1600 Earth days.
