1. The problem asks for the order of the element 20 in the group $\mathbb{Z}_{50}$. 2. Recall that $\mathbb{Z}_{50}$ is the group of integers modulo 50 under addition. 3. The order of an element $a$ in $\mathbb{Z}_n$ is the smallest positive integer $k$ such that $$k \cdot a \equiv 0 \pmod{n}.$$ 4. For $a=20$ and $n=50$, we want the smallest $k$ with $$k \times 20 \equiv 0 \pmod{50}.$$ 5. This means $50$ divides $20k$, or equivalently, $$50 \mid 20k.$$ 6. Simplify by dividing numerator and denominator by their greatest common divisor $\gcd(20,50) = 10$: $$\cancel{50} \mid \cancel{20}k \implies 5 \mid 2k.$$ 7. Since 5 divides $2k$, and 5 and 2 are coprime, 5 must divide $k$. 8. The smallest positive $k$ divisible by 5 is $k=5$. 9. Therefore, the order of 20 in $\mathbb{Z}_{50}$ is $5$. Final answer: 5 (option a).