For which value of $A$, does the equation $3m^2n = n^3 + A$ have a solution in natural numbers?

$ 
\textbf{(A)}\ 301
\qquad\textbf{(B)}\ 403
\qquad\textbf{(C)}\ 415
\qquad\textbf{(D)}\ 427
\qquad\textbf{(E)}\ 481
$
To determine the value of \( A \) for which the equation \( 3m^2n = n^3 + A \) has a solution in natural numbers, we can follow these steps:

1. **Rearrange the equation:**
   \[
   3m^2n = n^3 + A \implies A = 3m^2n - n^3
   \]

2. **Analyze the divisibility condition:**
   Since \( n \) is a natural number, \( n \) must divide \( A \). Therefore, \( A \) must be of the form \( n \cdot k \) for some integer \( k \).

3. **Consider the equation modulo 3:**
   \[
   3m^2n \equiv n^3 + A \pmod{3}
   \]
   Since \( 3m^2n \equiv 0 \pmod{3} \), we have:
   \[
   0 \equiv n^3 + A \pmod{3} \implies A \equiv -n^3 \pmod{3}
   \]

4. **Check the quadratic residue condition:**
   For \( A \) to be a valid solution, \( -n^3 \) must be a quadratic residue modulo 3. The quadratic residues modulo 3 are 0 and 1. Therefore, \( -n^3 \equiv 0 \) or \( -n^3 \equiv 1 \pmod{3} \).

5. **Evaluate the given options:**
   We need to check which of the given options satisfies the condition \( A \equiv -n^3 \pmod{3} \).

   - \( A = 301 \equiv 1 \pmod{3} \)
   - \( A = 403 \equiv 1 \pmod{3} \)
   - \( A = 415 \equiv 2 \pmod{3} \)
   - \( A = 427 \equiv 2 \pmod{3} \)
   - \( A = 481 \equiv 1 \pmod{3} \)

   Only \( A = 415 \) and \( A = 427 \) are not congruent to 1 modulo 3. 

6. **Factorize the remaining options:**
   - \( 415 = 5 \times 83 \)
   - \( 427 = 7 \times 61 \)

   We need to check if these factorizations can provide a solution in natural numbers for \( m \) and \( n \).

7. **Check for \( A = 415 \):**
   Let \( n = 5 \):
   \[
   3m^2 \cdot 5 = 5^3 + 415 \implies 15m^2 = 125 + 415 \implies 15m^2 = 540 \implies m^2 = 36 \implies m = 6
   \]
   This provides a valid solution in natural numbers.

8. **Check for \( A = 427 \):**
   Let \( n = 7 \):
   \[
   3m^2 \cdot 7 = 7^3 + 427 \implies 21m^2 = 343 + 427 \implies 21m^2 = 770 \implies m^2 = \frac{770}{21} \approx 36.67
   \]
   This does not provide a valid solution in natural numbers.

Therefore, the correct value of \( A \) is 415.

The final answer is \(\boxed{415}\).