A regular tetrahedron with $5$-inch edges weighs $2.5$ pounds. What is the weight in pounds of a similarly constructed regular tetrahedron that has $6$-inch edges? Express your answer as a decimal rounded to the nearest hundredth.
1. **Determine the volume ratio of the two tetrahedrons:**
   - The volume of a regular tetrahedron is proportional to the cube of its edge length. Therefore, if the edge lengths of two tetrahedrons are in the ratio $\frac{6}{5}$, then the ratio of their volumes is:
   \[
   \left(\frac{6}{5}\right)^3
   \]

2. **Calculate the volume ratio:**
   \[
   \left(\frac{6}{5}\right)^3 = \frac{6^3}{5^3} = \frac{216}{125}
   \]

3. **Simplify the volume ratio:**
   \[
   \frac{216}{125} = 1.728
   \]

4. **Determine the weight of the larger tetrahedron:**
   - Since the weight of the tetrahedron is proportional to its volume, the weight of the larger tetrahedron is:
   \[
   2.5 \times 1.728
   \]

5. **Calculate the weight:**
   \[
   2.5 \times 1.728 = 4.32
   \]

The final answer is $\boxed{4.32}$