What is the $33$-rd number after the decimal point of $(\sqrt {10} + 3)^{2001}$?

$ 
\textbf{(A)}\ 0
\qquad\textbf{(B)}\ 1
\qquad\textbf{(C)}\ 2
\qquad\textbf{(D)}\ 4
\qquad\textbf{(E)}\ 8
$
1. We start by considering the expression \((\sqrt{10} + 3)^{2001}\). To find the 33rd digit after the decimal point, we need to understand the behavior of this expression.

2. By the Binomial Theorem, we can write:
   \[
   (\sqrt{10} + 3)^{2001} = \sum_{k=0}^{2001} \binom{2001}{k} (\sqrt{10})^k \cdot 3^{2001-k}
   \]
   Similarly, we can write:
   \[
   (\sqrt{10} - 3)^{2001} = \sum_{k=0}^{2001} \binom{2001}{k} (\sqrt{10})^k \cdot (-3)^{2001-k}
   \]

3. Let \(N\) be a natural number such that:
   \[
   N = (\sqrt{10} + 3)^{2001} - (\sqrt{10} - 3)^{2001}
   \]
   This \(N\) is an integer because the irrational parts cancel out when the two expressions are subtracted.

4. Now, consider the term \((\sqrt{10} - 3)^{2001}\). Since \(\sqrt{10} \approx 3.162\), we have:
   \[
   \sqrt{10} - 3 \approx 0.162
   \]
   Therefore:
   \[
   0 < \sqrt{10} - 3 < 1
   \]

5. We need to estimate \((\sqrt{10} - 3)^{2001}\). Since \(\sqrt{10} - 3 < \frac{1}{6}\), we have:
   \[
   (\sqrt{10} - 3)^{2001} < \left(\frac{1}{6}\right)^{2001}
   \]

6. To understand the magnitude of \(\left(\frac{1}{6}\right)^{2001}\), we note that:
   \[
   \left(\frac{1}{6}\right)^{2001} = \frac{1}{6^{2001}}
   \]
   Since \(6^{2001}\) is a very large number, \(\frac{1}{6^{2001}}\) is extremely small.

7. We can further approximate:
   \[
   \left(\frac{1}{6}\right)^{2001} < \frac{1}{10^{2001 \log_{10}(6)}}
   \]
   Since \(\log_{10}(6) \approx 0.778\), we have:
   \[
   2001 \log_{10}(6) \approx 1556.778
   \]
   Therefore:
   \[
   \left(\frac{1}{6}\right)^{2001} < \frac{1}{10^{1556.778}}
   \]

8. This means that \((\sqrt{10} - 3)^{2001}\) is less than \(10^{-1556.778}\), which implies that it has at least 1556 zeros after the decimal point.

9. Since \((\sqrt{10} - 3)^{2001}\) is extremely small, adding it to \(N\) (an integer) does not affect the first 1556 digits after the decimal point of \((\sqrt{10} + 3)^{2001}\).

10. Therefore, the 33rd digit after the decimal point of \((\sqrt{10} + 3)^{2001}\) is \(0\).

The final answer is \(\boxed{0}\).