Let's re-evaluate the problem:

To find \( \log_3 (9 \sqrt{81}) \):

1. Simplify \( 9 \sqrt{81} \):
- \( 9 = 3^2 \)
- \( \sqrt{81} = \sqrt{3^4} = 3^2 \)
- Therefore, \( 9 \sqrt{81} = 3^2 \cdot 3^2 = 3^{2+2} = 3^4 \)

2. Apply the logarithm:
- \( \log_3 (3^4) = 4 \cdot \log_3 (3) \)
- Since \( \log_3 (3) = 1 \), this simplifies to \( 4 \cdot 1 = 4 \)


Let's correct the base expression and calculation:

- Rewrite \( 9 \sqrt{81} \) in terms of powers of 3:
- \( 9 = 3^2 \)
- \( \sqrt{81} = \sqrt{3^4} = 3^2 \)
- Therefore, \( 9 \sqrt{81} = 3^2 \cdot 3^2 = 3^4 \)

So:

\[
\log_3 (9 \sqrt{81}) = \log_3 (3^4) = 4
\]