Planetary mystery

The planet's period is $8$ Earth Years.

Actually, Kepler's Laws can be applied, because a circle is considered to be a special case of an ellipse. So, the equation is:

$T = \dfrac{2\pi a^{3/2}}{\sqrt{Gm_S}}$

where $a$ is the length of the semi-major axis, $G = 6.67 \times 10^{-11} \frac{Nm^2}{kg^2}$ is the gravitational constant, and $m_S$ is the mass of the Sun.

In the case of circular orbits, however, the radius $r$ will now be equal to the length of the semi-major axis $a$ , that is: $r = a$ .

$T = \dfrac{2\pi r^{3/2}}{\sqrt{Gm_S}}$

Knowing that the radius of the planet's orbit $r_P$ is $4$ times the radius of Earth's orbit $r_E$ , that is: $r_P = 4r_E$ :

$T_E = \dfrac{2\pi r_E^{\frac{3}{2}}}{\sqrt{Gm_S}}$

$T_P = \dfrac{2\pi r_P^{\frac{3}{2}}}{\sqrt{Gm_S}} = \dfrac{2\pi {(4r_E)}^{\frac{3}{2}}}{\sqrt{Gm_S}}$

$T_P = 8 \times \dfrac{2\pi r_E^{\frac{3}{2}}}{\sqrt{Gm_S}} = 8T_E$

The planet's period is $\boxed{8}$ times the period of the planet Earth.

T^2 directly proportional to a^3 (T1/T2)^2=(a1/a2)^3 T1/1=(4r/r)^3/2 T1=8 years