In an acute triangle $ ABC$, let $ D$ be a point on $ \left[AC\right]$  and $ E$ be a point on $ \left[AB\right]$ such that $ \angle ADB\equal{}\angle AEC\equal{}90{}^\circ$. If perimeter of triangle $ AED$ is 9, circumradius of  $ AED$ is $ \frac{9}{5}$ and perimeter of triangle $ ABC$ is 15, then $ \left|BC\right|$ is

$\textbf{(A)}\ 5 \qquad\textbf{(B)}\  \frac{24}{5} \qquad\textbf{(C)}\ 6 \qquad\textbf{(D)}\ 8 \qquad\textbf{(E)}\  \frac{48}{5}$
1. **Identify the given information and relationships:**
   - Triangle \( ABC \) is acute.
   - Points \( D \) and \( E \) are on segments \( [AC] \) and \( [AB] \) respectively, such that \( \angle ADB = \angle AEC = 90^\circ \).
   - The perimeter of triangle \( AED \) is 9.
   - The circumradius of triangle \( AED \) is \( \frac{9}{5} \).
   - The perimeter of triangle \( ABC \) is 15.

2. **Establish the cyclic nature of quadrilateral \( BCDE \):**
   - Since \( \angle ADB = \angle AEC = 90^\circ \), quadrilateral \( BCDE \) is cyclic.
   - This implies that \( \angle ABC = \angle ADE \) and \( \angle ACB = \angle AED \).

3. **Use similarity of triangles \( ADE \) and \( ABC \):**
   - Since \( \angle ADE = \angle ABC \) and \( \angle AED = \angle ACB \), triangles \( ADE \) and \( ABC \) are similar.
   - The similarity ratio is given by the ratio of their perimeters: \( \frac{\text{Perimeter of } ADE}{\text{Perimeter of } ABC} = \frac{9}{15} = \frac{3}{5} \).

4. **Determine the ratio of corresponding sides:**
   - Let \( AD \) and \( AB \) be corresponding sides in triangles \( ADE \) and \( ABC \) respectively.
   - Therefore, \( \frac{AD}{AB} = \frac{3}{5} \).

5. **Calculate the circumradius of triangle \( ABC \):**
   - The circumradius of similar triangles is also in the same ratio as their sides.
   - Given the circumradius of \( AED \) is \( \frac{9}{5} \), the circumradius \( R \) of \( ABC \) can be found using the ratio:
     \[
     \frac{\frac{9}{5}}{R} = \frac{3}{5} \implies R = \frac{\frac{9}{5} \cdot 5}{3} = 3
     \]

6. **Use the circumradius to find \( BC \):**
   - In triangle \( ABC \), the formula for the circumradius \( R \) is given by:
     \[
     R = \frac{a}{2 \sin A}
     \]
     where \( a = BC \) and \( \sin A \) is the sine of angle \( A \).
   - From the similarity ratio, we know \( \sin A = \frac{4}{5} \) (since \( \sin A \) is preserved in similar triangles).
   - Therefore:
     \[
     BC = 2R \sin A = 2 \cdot 3 \cdot \frac{4}{5} = \frac{24}{5}
     \]

The final answer is \(\boxed{\frac{24}{5}}\).