Answer
See below.
Work Step by Step
The odds for $E$: $\frac{P(E)}{P(\text{not E})}$. The odds against $E$: $\frac{P(\text{not E})}{P(E)}$
The total number of outcomes in this sample: $2000$, the number of outcomes satisfying the requirements in this sample: $20$, thus the odds for: $\frac{20}{2000-20}=\frac{20}{1980}=\frac{1}{99}$ and thus the odds against: $\frac{1}{\text{odds for E}}=\frac{99}{1}=99$
