The distance from the Sun to Mercury is 0.4 AU. Determine the sidereal period for this planet.

To solve the problem, you need to use Kepler’s third law.
Kepler’s Law: The squares of the orbital periods of planets around the Sun are referred to as cubes of the semi-major axes of their elliptical orbits.
T * m ^ 2 / Tz ^ 2 = am ^ 3 / az ^ 3
T * m = √ (Tz ^ 2 * am ^ 3 / az ^ 3) = √ (1 ^ 2 * 0.4 ^ 3/1 ^ 3) = 0.25 Earth years