The length of a cycle is the number of vertices in the cycle (which is equal to the number of edges in the cycle). Some examples are given below:
Suppose we have a graph $G$ with $n \geq 2$ vertices and minimum degree $3$. Let $C$ be the shortest cycle in $G$, and let $x$ be a vertex in $C$. (Note: there exists at least one cycle in $G$, otherwise $G$ would be a forest, and non-empty forests have vertices of degree $0$ or $1$.)
Two slightly different situations can arise, depending on whether $C$ is an odd-length cycle or an even-length cycle.
Case I: $C$ is a $t$-cycle, where $t$ is odd.
To find the upper bound on $t$, we look at the neighbours of $x$, then their neighbours, and so on, until we first hit a cycle. This will occur at depth $\frac{t-1}{2}$, since $t$ is odd.
The situations is as depicted below, where the cycle in bold represents $C$.
Every vertex on level $i \in \{1,2,\ldots,\frac{t-1}{2}-1\}$ is connected to at least $2$ vertices on level $i+1$ and exactly $1$ vertex on level $i-1$. Any other situation would either imply (a) we have not accounted for all vertices at a given level or (b) there exists a cycle shorter than $C$. Therefore
\begin{align*}
n &\geq 1+\sum_{i=0}^{\frac{t-1}{2}-1} 3 \cdot 2^i \\
&=1+3 \cdot 2^{\frac{t-1}{2}} & \text{using the identity } \sum_{k=0}^a 2^a=2^{a+1}.
\end{align*}
Case II: $C$ is a $t$-cycle, where $t$ is even.
In this situation, the picture looks like the following:
And we can instead deduce
\begin{align*}
n &\geq 3+\sum_{i=0}^{\frac{t}{2}-2} 3 \cdot 2^i \\
&=3+3 \cdot 2^{\frac{t}{2}-1} & \text{using the identity } \sum_{k=0}^a 2^a=2^{a+1}.
\end{align*}
Hence we have:
Lemma: $n > 2^{\frac{t}{2}}$ for all $n$.
Proof: If $t$ is odd, then $$n \geq 1+3 \cdot 2^{\frac{t-1}{2}} = 1+\frac{3}{\sqrt{2}} \cdot 2^{\frac{t}{2}} \geq 2^{\frac{t}{2}}.$$ If $t$ is even, then $$n \geq 3+3 \cdot 2^{\frac{t}{2}-1} = 3+\frac{3}{2} \cdot 2^{\frac{t}{2}} \geq 2^{\frac{t}{2}}.$$ [End of proof.]
So, if $t>2 \lceil \log_2 n \rceil$, then, by the lemma, we have $$n > 2^{\log_2 n} = n$$ giving a contradiction. So finally, we can conclude:
Theorem: If $G$ is a graph with minimum degree $\geq 3$, then there exists a cycle of length at most $2 \lceil \log_2 n \rceil$.
It is worth noting that the bound can sometimes be achieved. For example, this is $K_{3,3}$:
which has $n=6$ vertices and the shortest cycle is a $2 \lceil \log_2 n \rceil$-cycle. It's not going to be "perfect" (since we rounded off in the lemma above), but it will be asymptotically the best possible.
Jernej is correct, $|V(G)| = n$ and $|E(G)|=2n-3$ does not imply that $G$ is connected. Counter example: Consider $n = 6$, then $2n-3=9$. You can construct a graph consisting of a single isolated node and an almost completely connected component with 5 vertices. This is possible because $|E(K_5)| = 10 > 9$.
WLOG let $G$ have $k$ connected components $c_i$, $i=1,\dots,k$. Let $c_j$ be such that $|V(c_j)| \geq 4$ and $|E(c_j)| \geq 2\cdot|V(c_j)| - 3$. Such a $c_j$ must exist because $G$ could not have $2n-3$ edges otherwise. Now consider $T$, the spanning tree of $c_j$. It consists of $|V(c_j)| - 1$ edges, so there are at least $|V(c_j)| - 2$ edges left. Each of these edges creates a cycle of length between 3 and $|V(c_j)|$ when joined with $T$. Since there are $|V(c_j)| - 2$ edges and $|V(c_j)| - 2$ possible distinct cycle lengths this is not yet enough to apply the pigeon hole principle. However, observe that when there is a cycle of length $|V(c_j)|$ every additional edge will create at least two new cycles. Hence either there will be a repeated cycle length when choosing $|V(c_j)|-2$ lengths with $|V(c_j)|-2$ edges, or if not, then there must be a cycle of length $|V(c_j)|$ and so there will be $2(|V(c_j)|-3)+1$ cycles. $2(|V(c_j)|-3)+1$ > $|V(c_j)|-2$ for all $|V(c_j)| > 3$ so the pigeon hole principle can be applied in this case. $\quad\square$
Best Answer
This follows from the fact that every graph with $\chi(G) \geq 3$ has a cycle of length at least $\chi(G)$ (see Corollary 3 in this answer).
To solve the problem at hand, assume that $\alpha(G) < \sqrt{n}$. Then $\chi(G) \geq \frac{n}{\alpha(G)} > \sqrt{n} > 2$, so one has $\chi(G) \geq 3$, and it follows that $G$ has a cycle of length at least $\sqrt{n}$.