Let $X$ be the total number of cycles,
$X_k$ the number of cycles of length $k$ in $G$, and
$Y_k$ the number of cycles of length $k$ in a random graph of size $k$.
Note that:
- ${\sf E}[X] = \sum_{k=2}^n {\sf E}[X_k]$.
- ${\sf E}[X_k] = {n \choose k} {\sf E}[Y_k]$.
- ${\sf E}[Y_k] = (k!/k)p^k = (k-1)!p^k$.
The last one is because a cycle of length $k$
is a permutation of numbers from $1$ to $k$
but the first vertex does not matter (a cycle does not have a first vertex).
The probability of each edge being present is $p$,
so the probability of $k$ edges being present is $p^k$.
We have:
\begin{align*}
{\sf E}[X] &= \sum_{k=2}^n E[X_k] \\
&= \sum_{k=2}^n {n \choose k} E[Y_k] \\
&= \sum_{k=2}^n {n \choose k} (k-1)! p^k \\
&= \sum_{k=2}^n \frac{n!}{k!(n-k)!} (k-1)! p^k \\
&= \sum_{k=2}^n \frac{n!}{(n-k)!k} p^k \\
&= \sum_{k=2}^n \frac{n\ldots (n-k+1)}{k} p^k \\
\end{align*}
For a simple lower bound
we can drop the first half of the sum:
\begin{align*}
&\geq \sum_{k=n/2}^{n} \frac{n\ldots (n-k+1)}{k} p^k \\
&\geq \sum_{k=n/2}^{n} \frac{(n/2)^{k}}{n} p^k \\
&\geq \frac{1}{n}\sum_{k=n/2}^{n} (np/2)^{k} \\
&= \frac{(\frac{np}{2})^{n+1}-(\frac{np}{2})^{\frac{n}{2}}}{\frac{np}{2}-1}
\end{align*}