A positive real number $ A $ rounds to 20, and another positive real number $ B $ rounds to 23. What is the largest possible value of the largest integer not exceeding the value of $ \frac{100A}{B}? $

$\textbf{(A) } 91 \qquad \textbf{(B) } 89 \qquad \textbf{(C) } 88 \qquad \textbf{(D) } 87 \qquad \textbf{(E) } \text{none of the above}$
1. **Identify the range of values for \( A \) and \( B \):**
   - Since \( A \) rounds to 20, \( A \) must be in the interval \( [19.5, 20.5) \).
   - Since \( B \) rounds to 23, \( B \) must be in the interval \( [22.5, 23.5) \).

2. **Determine the maximum value of \( \frac{100A}{B} \):**
   - To maximize \( \frac{100A}{B} \), we need to maximize \( A \) and minimize \( B \).
   - The maximum value of \( A \) is just under 20.5, i.e., \( A \approx 20.499\ldots \).
   - The minimum value of \( B \) is 22.5.

3. **Calculate \( \frac{100A}{B} \) using the extreme values:**
   \[
   \frac{100A}{B} \approx \frac{100 \times 20.499\ldots}{22.5}
   \]
   \[
   \frac{100 \times 20.499\ldots}{22.5} = \frac{2049.9\ldots}{22.5} \approx 91.107\ldots
   \]

4. **Find the largest integer not exceeding \( \frac{100A}{B} \):**
   - The largest integer not exceeding \( 91.107\ldots \) is 91.

Conclusion:
\[
\boxed{91}
\]