For what digit $A$ is the numeral $1AA$ a perfect square in base-$5$ and a perfect cube in base-$6$?

$\text{(A) }0\qquad\text{(B) }1\qquad\text{(C) }2\qquad\text{(D) }3\qquad\text{(E) }4$
To solve the problem, we need to find the digit \( A \) such that the numeral \( 1AA \) is both a perfect square in base-5 and a perfect cube in base-6. 

1. **Convert \( 1AA_5 \) to base-10:**
   \[
   1AA_5 = 1 \cdot 5^2 + A \cdot 5 + A = 25 + 6A
   \]
   We need \( 25 + 6A \) to be a perfect square, i.e., \( 25 + 6A = x^2 \) for some integer \( x \).

2. **Convert \( 1AA_6 \) to base-10:**
   \[
   1AA_6 = 1 \cdot 6^2 + A \cdot 6 + A = 36 + 7A
   \]
   We need \( 36 + 7A \) to be a perfect cube, i.e., \( 36 + 7A = y^3 \) for some integer \( y \).

3. **Determine the possible values of \( A \):**
   Since \( A \) is a digit in base-5 and base-6, it must be in the set \(\{0, 1, 2, 3, 4\}\).

4. **Check each value of \( A \) to satisfy both conditions:**
   - For \( A = 0 \):
     \[
     25 + 6 \cdot 0 = 25 \quad \text{(perfect square: } 5^2)
     \]
     \[
     36 + 7 \cdot 0 = 36 \quad \text{(not a perfect cube)}
     \]
   - For \( A = 1 \):
     \[
     25 + 6 \cdot 1 = 31 \quad \text{(not a perfect square)}
     \]
   - For \( A = 2 \):
     \[
     25 + 6 \cdot 2 = 37 \quad \text{(not a perfect square)}
     \]
   - For \( A = 3 \):
     \[
     25 + 6 \cdot 3 = 43 \quad \text{(not a perfect square)}
     \]
   - For \( A = 4 \):
     \[
     25 + 6 \cdot 4 = 49 \quad \text{(perfect square: } 7^2)
     \]
     \[
     36 + 7 \cdot 4 = 64 \quad \text{(perfect cube: } 4^3)
     \]

Since \( A = 4 \) satisfies both conditions, we conclude that \( A = 4 \) is the correct digit.

The final answer is \( \boxed{4} \).