What is the sum of $3+3^2+3^{2^2} + 3^{2^3} + \dots + 3^{2^{2006}}$ in $\mod 11$?

$ 
\textbf{(A)}\ 0
\qquad\textbf{(B)}\ 1
\qquad\textbf{(C)}\ 2
\qquad\textbf{(D)}\ 5
\qquad\textbf{(E)}\ 10
$
1. **Identify the sequence pattern:**
   We need to find the sum of the series \(3 + 3^2 + 3^{2^2} + 3^{2^3} + \dots + 3^{2^{2006}}\) modulo 11. First, we observe the behavior of powers of 3 modulo 11.

2. **Calculate the first few powers of 3 modulo 11:**
   \[
   3^1 \equiv 3 \mod 11
   \]
   \[
   3^2 \equiv 9 \mod 11
   \]
   \[
   3^4 \equiv (3^2)^2 \equiv 9^2 \equiv 81 \equiv 4 \mod 11
   \]
   \[
   3^8 \equiv (3^4)^2 \equiv 4^2 \equiv 16 \equiv 5 \mod 11
   \]
   \[
   3^{16} \equiv (3^8)^2 \equiv 5^2 \equiv 25 \equiv 3 \mod 11
   \]

3. **Identify the repeating cycle:**
   We observe that the sequence of powers of 3 modulo 11 repeats every 4 terms:
   \[
   3, 9, 4, 5, 3, 9, 4, 5, \ldots
   \]

4. **Determine the number of complete cycles:**
   Since the sequence repeats every 4 terms, we need to determine how many complete cycles fit into the first 2006 terms. We have:
   \[
   2006 \div 4 = 501 \text{ complete cycles with a remainder of } 2
   \]

5. **Sum the terms in one complete cycle:**
   The sum of one complete cycle is:
   \[
   3 + 9 + 4 + 5 = 21 \equiv 10 \mod 11
   \]

6. **Sum the terms of the remaining part:**
   The remaining part consists of the first 2 terms of the next cycle:
   \[
   3 + 9 = 12 \equiv 1 \mod 11
   \]

7. **Calculate the total sum:**
   The total sum is the sum of 501 complete cycles plus the sum of the remaining part:
   \[
   501 \times 10 + 1 \equiv 5010 + 1 \equiv 6 \times 10 + 1 \equiv 60 + 1 \equiv 61 \equiv 6 \mod 11
   \]

The final answer is \(\boxed{6}\)