Answer:
4.6 years
Step-by-step explanation:
This is solved using Kepler's third law which says:

Where
T = Orbital period of the planet (in seconds)
a = Distance from the star (in meters)
G = Gravitational constant
M = Mass of the parent star (in kg)
From the information given



We put this into Kepler's law and get:

This when converted to years is 4.6 years.