Solution:

To find the time taken by a planet at a distance of \( 540 \times 10^{10} \) m to revolve around the Sun, we can use Kepler's Third Law:

\[
\frac{T_1^2}{T_2^2} = \frac{r_1^3}{r_2^3}
\]

 Given:
- Earth:
- \( T_1 = 1 \) year
- \( r_1 = 15 \times 10^{10} \) m
- Planet:
- \( r_2 = 540 \times 10^{10} \) m

 Calculation:
1. Set up the ratio:
\[
\frac{1^2}{T_2^2} = \left( \frac{15}{540} \right)^3
\]
\[
\frac{15}{540} = \frac{1}{36} \quad \Rightarrow \quad \left( \frac{1}{36} \right)^3 = \frac{1}{46656}
\]

2. Solve for \( T_2^2 \):
\[
T_2^2 = 46656 \quad \Rightarrow \quad T_2 = \sqrt{46656} = 216 \text{ years}
\]

The time taken by the planet to revolve around the Sun once is approximately 216 years.