Elements of a prime order

An element $a$ has order $n$ then $a^k$ has order $|a^k|=\dfrac{n}{gcd(n,k)}$

So here $|a|=5$ so $\dfrac{5}{gcd(5,k)}=5 \implies gcd(5,k)=1 \implies k=1,2,3,4$

There are four elements $a,a^2,a^3,a^4$ has order $5$

We know from Sylow's Theorem that $n_{5} \equiv 1 \mod 5$ . This gives that $n_{5} \in \{1 + 5n | n \in \mathbb{N}\}$ . As $|G| = 20$ , we get that: $n_{5} >1 \implies |G| \geq 30$ . Therefore $n_{5} = 1$ . Then for the sole $P \in Syl_{5}(G)$ , we have that: $\forall g \in P$ , the order of $g$ divides the order of $P$ , which is $5$ . There exists a single identity element of order one in $P$ , and thus all other elements in $P$ are of order $5$ . Thus we get the answer of $4$ elements existing in $G$ of order $5$ .