Not 100% sure what the question is but i’m assuming the answer is C. spectrum
To solve this problem we will use Kepler's third law of planetary motion. Formula for that law is:

Where:
T = orbital period
G = universal gravitational constant
M = mass of star
a = distance from planet to star
By rearranging this formula we get:

For Earth we have:

For Wasp-32b we have:

or
We can see that right sides are same. Left sides must be same. We can use this to solve for distance.

![\frac{365^{2}}{1.1* 2.7^{2}} = \frac{1}{a^{3} } \\ \\ 2.7^{2} =365^{2}*a^{3} \\ \\ a^{3} =\frac{2.7^{2} }{365^{2}} \\ \\ a= \sqrt[3]{\frac{2.7^{2} }{365^{2}} } \\ \\ \\ a=0.038AU](https://tex.z-dn.net/?f=%5Cfrac%7B365%5E%7B2%7D%7D%7B1.1%2A%202.7%5E%7B2%7D%7D%20%3D%20%5Cfrac%7B1%7D%7Ba%5E%7B3%7D%20%7D%20%5C%5C%20%5C%5C%202.7%5E%7B2%7D%20%3D365%5E%7B2%7D%2Aa%5E%7B3%7D%20%5C%5C%20%5C%5C%20a%5E%7B3%7D%20%3D%5Cfrac%7B2.7%5E%7B2%7D%20%7D%7B365%5E%7B2%7D%7D%20%5C%5C%20%5C%5C%20a%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7B2.7%5E%7B2%7D%20%7D%7B365%5E%7B2%7D%7D%20%7D%20%5C%5C%20%5C%5C%20%5C%5C%20a%3D0.038AU)
Wasp-32b is at distance of 0.038AU.
Your answer is 3,721 feet.