The sum of the first 2011 terms of a geometric series is 200. The sum of the first 4022 terms of the same series is 380. Find the sum of the first 6033 terms of the series.
1. **Identify the given information and the geometric series formula:**
   - The sum of the first 2011 terms of a geometric series is 200.
   - The sum of the first 4022 terms of the same series is 380.
   - We need to find the sum of the first 6033 terms of the series.

   The sum \( S_n \) of the first \( n \) terms of a geometric series is given by:
   \[
   S_n = \frac{a(1 - r^n)}{1 - r}
   \]
   where \( a \) is the first term and \( r \) is the common ratio.

2. **Set up the equations using the given sums:**
   \[
   \frac{a(1 - r^{2011})}{1 - r} = 200 \quad \text{(1)}
   \]
   \[
   \frac{a(1 - r^{4022})}{1 - r} = 380 \quad \text{(2)}
   \]

3. **Express \( 1 - r^{4022} \) in terms of \( 1 - r^{2011} \):**
   Notice that:
   \[
   1 - r^{4022} = (1 - r^{2011})(1 + r^{2011})
   \]

4. **Divide equation (2) by equation (1):**
   \[
   \frac{\frac{a(1 - r^{4022})}{1 - r}}{\frac{a(1 - r^{2011})}{1 - r}} = \frac{380}{200}
   \]
   Simplifying, we get:
   \[
   \frac{1 - r^{4022}}{1 - r^{2011}} = \frac{19}{10}
   \]

5. **Substitute \( 1 - r^{4022} = (1 - r^{2011})(1 + r^{2011}) \) into the equation:**
   \[
   \frac{(1 - r^{2011})(1 + r^{2011})}{1 - r^{2011}} = \frac{19}{10}
   \]
   Simplifying, we get:
   \[
   1 + r^{2011} = \frac{19}{10}
   \]

6. **Solve for \( r^{2011} \):**
   \[
   10 + 10r^{2011} = 19
   \]
   \[
   10r^{2011} = 9
   \]
   \[
   r^{2011} = \frac{9}{10}
   \]

7. **Substitute \( r^{2011} = \frac{9}{10} \) back into equation (1) to find \( \frac{a}{1 - r} \):**
   \[
   \frac{a(1 - \frac{9}{10})}{1 - r} = 200
   \]
   \[
   \frac{a \cdot \frac{1}{10}}{1 - r} = 200
   \]
   \[
   \frac{a}{1 - r} = 2000
   \]

8. **Find the sum of the first 6033 terms:**
   \[
   S_{6033} = \frac{a(1 - r^{6033})}{1 - r}
   \]
   Notice that \( r^{6033} = (r^{2011})^3 \):
   \[
   r^{6033} = \left(\frac{9}{10}\right)^3 = \frac{729}{1000}
   \]
   Therefore:
   \[
   S_{6033} = \frac{a}{1 - r} \cdot (1 - r^{6033})
   \]
   \[
   S_{6033} = 2000 \cdot \left(1 - \frac{729}{1000}\right)
   \]
   \[
   S_{6033} = 2000 \cdot \frac{271}{1000}
   \]
   \[
   S_{6033} = 2000 \cdot 0.271 = 542
   \]

The final answer is \(\boxed{542}\).