One law of planetary motion states that the square of a planets orbital period, T, is proportional to the cube of the radius, r. This relationship can be written as an equation: T2=r3.

To solve for the distance of the planet from the Sun in Astronomical Units (AU), we can use Kepler's Third Law of Planetary Motion, which states that

\[ T^2 \propto r^3 \]

We can express this as:

\[ T^2 = k \cdot r^3 \]

where \( k \) is a constant that can be determined based on Earth’s orbit, where \( T = 1 \) year and \( r = 1 \) AU. For Earth:

\[ 1^2 = k \cdot 1^3 \]
\[ k = 1 \]

Now we can use this relationship to determine \( r \) for the given orbital period \( T = 29.46 \) years:

1. Calculate \( T^2 \):

\[ T^2 = (29.46)^2 = 869.0516 \]

2. Set that equal to \( r^3 \):

\[ r^3 = 869.0516 \]

3. Now solve for \( r \) by taking the cube root:

\[ r = \sqrt[3]{869.0516} \]

Calculating the cube root:

\[ r \approx 9.54 \, \text{AU} \]

Thus, the planet is approximately 9.54 AU away from the Sun.

Therefore, the correct response is:

**9.54 AU**