7. kepler’s third law of orbital motion states that you can approximate the period p (in earth years) it takes a planet to complete one orbit of the sun using the function , where d is the distance (in astronomical units, au) from the planet to the sun. how many earth years would it take for a planet that is 6.76 au from the sun? 15.23
17.58
154.46
3.58

In your question, this function is supposed to be in the little space after the word "function":

(square of the orbital period) = k · (cube of the distance from the sun)

' k ' is the same number for every solar-system object

To answer the question, we first have to find out what ' k ' is.  Since it's the same number for every planet in our solar system, we can find ' k ' for the Earth, and then use it for the mystery planet.

Period = 1 Earth year

Distance = 1 Astronomical unit

P² = k · D³

1² = k · 1³

k = 1  

Now for the mystery planet:  

P² = k · D³

P² = 1 · (6.76)³

P² = 308.9 Earth years²

P = √308.9

P = 17.58 Earth years