1. **State the problem:** Calculate the sum of the first 6 terms of a geometric series with first term $a_1 = -21$ and common ratio $r = -\frac{2}{3}$. 2. **Formula for sum of first n terms of geometric series:** $$S_n = a_1 \frac{1 - r^n}{1 - r}$$ 3. **Substitute given values:** $$S_6 = -21 \times \frac{1 - \left(-\frac{2}{3}\right)^6}{1 - \left(-\frac{2}{3}\right)}$$ 4. **Calculate $r^6$:** $$\left(-\frac{2}{3}\right)^6 = \left(\frac{2}{3}\right)^6 = \frac{2^6}{3^6} = \frac{64}{729}$$ 5. **Evaluate numerator:** $$1 - \frac{64}{729} = \frac{729}{729} - \frac{64}{729} = \frac{665}{729}$$ 6. **Evaluate denominator:** $$1 - \left(-\frac{2}{3}\right) = 1 + \frac{2}{3} = \frac{3}{3} + \frac{2}{3} = \frac{5}{3}$$ 7. **Substitute back:** $$S_6 = -21 \times \frac{\frac{665}{729}}{\frac{5}{3}}$$ 8. **Divide fractions:** $$\frac{\frac{665}{729}}{\frac{5}{3}} = \frac{665}{729} \times \frac{3}{5} = \frac{665 \times 3}{729 \times 5} = \frac{1995}{3645}$$ 9. **Simplify fraction:** $$\frac{1995}{3645} = \frac{\cancel{1995}^{\times 3 \times 5}}{\cancel{3645}^{\times 3 \times 5}} = \frac{1995 \div 15}{3645 \div 15} = \frac{133}{243}$$ 10. **Multiply by -21:** $$S_6 = -21 \times \frac{133}{243} = \frac{-21 \times 133}{243} = \frac{-2793}{243}$$ **Final answer:** $$\boxed{-\frac{2793}{243}}$$ which corresponds to option c.