Determine the smallest positive integer, $n$, which satisfies the equation $n^3+2n^2 = b$, where $b$ is the square of an odd integer.
1. We start with the given equation:
   \[
   n^3 + 2n^2 = b
   \]
   where \( b \) is the square of an odd integer.

2. We can factor the left-hand side:
   \[
   n^3 + 2n^2 = n^2(n + 2)
   \]
   Let \( b = k^2 \) for some odd integer \( k \). Thus, we have:
   \[
   n^2(n + 2) = k^2
   \]

3. Since \( k \) is an odd integer, \( k^2 \) is also odd. For \( n^2(n + 2) \) to be odd, \( n \) must be odd because the product of an even number and an odd number is even, and the product of two odd numbers is odd.

4. Let \( n \) be an odd integer. We need to find the smallest positive odd integer \( n \) such that \( n^2(n + 2) \) is a perfect square.

5. We test small odd integers for \( n \):

   - For \( n = 1 \):
     \[
     n^2(n + 2) = 1^2(1 + 2) = 1 \cdot 3 = 3 \quad (\text{not a perfect square})
     \]

   - For \( n = 3 \):
     \[
     n^2(n + 2) = 3^2(3 + 2) = 9 \cdot 5 = 45 \quad (\text{not a perfect square})
     \]

   - For \( n = 5 \):
     \[
     n^2(n + 2) = 5^2(5 + 2) = 25 \cdot 7 = 175 \quad (\text{not a perfect square})
     \]

   - For \( n = 7 \):
     \[
     n^2(n + 2) = 7^2(7 + 2) = 49 \cdot 9 = 441 = 21^2 \quad (\text{a perfect square})
     \]

6. We find that \( n = 7 \) satisfies the equation \( n^2(n + 2) = k^2 \) where \( k = 21 \), an odd integer.

Conclusion:
\[
\boxed{7}
\]