Nondeterministic Logarithmic-space in directed graph
To show that the problem is in NL: guess a directed edge $(u,v)$ of the graph that is part of a cycle, and check (using the connectivity algorithm as a black-box, or by guessing each step of a walk) whether $v$ is connected to $u$ in $G$.

To show that the problem is NL-Hard create a new graph $G'$ as follows:


For each  vertex $v$ of $G$ add $n$ vertices $v^{(0)},\dots,v^{(n-1)}$ to $G'$.
For each edge $(u,v)$ of $G$, and for each $i=0,\dots,n-2$, add the edge $(u^{(i)}, v^{(i+1)})$ to $G'$.
For each $i=0,\dots,n-2$, add the edge $(t^{(i)}, t^{(i+1)})$ to $G'$.


It is easy to see that $G'$ is acyclic and that $s^{(0)}$ and $t^{(n-1)}$ are connected in $G'$ iff $s$ and $t$ are connected in $G$. 

Now, consider the graph $G''$ obtained by adding the edge $e=(t^{(n-1)}, s^{(0)})$ to $G'$.

If $s^{(0)}$ and $t^{(n-1)}$ were connected by a path $P$ in $G'$, $G''$ contains the cycle $C = P + e$. The opposite direction is also true: if there is a cycle $C$ in $G''$ then $C$ must include $e$ and this means that of $C-e$ is a path between $s^{(0)}$ and $t^{(n-1)}$ in $G'$.