Let $c > 0$ and set $p := \frac{c}{n^{2/3}}$. Use Janson's Inequality to find a function $q(c): \mathbb{R}_{>0} \rightarrow (0,1)$ such that $$\mathbb{P}\left[ \text{$G(n,p)$ contains no clique of size $4$} \right] \overset{n \rightarrow \infty}\longrightarrow q(c).$$
Remark: For the version of Janson's inequality that I know see this picture from "Probabilistic Methods in Combinatorics" by Yufei Zhao below).
I am not sure whether the $4$ plays a significant role here, so let's set the RV $X$ to be the number of $k$-cliques and let $A_i$ be the event that the $i$-th of the $n \choose k$ $k$-subsets is a clique. Then expected value $\mu$ can be computed by counting the number of $k$-subsets times the probability that a $k$-subset is a clique:
$$\mu = \mathbb{E}[X] = {n \choose k} p^{- {k \choose 2}}.$$
However, concerning $\Delta$ I use the formula
\begin{align} \Delta &= \sum_{i \sim j} \mathbb{P}\left[ A_i \cap A_j \right], \\ \end{align}
but I do not see how to compute $\mathbb{P} \left[ A_i \cap A_j \right]$ when $A_i$ and $A_j$ are not independent.
Could you please give me a hint?
I'll show the argument for $4$ and you should be able to extend to general $k$. Let us consider the different cases.
$i$-th and $j$-th subsets have no common node: There are $\binom{n}{4} \binom{n-4}{4}$ subsets. In this case, $A_i$ and $A_j$ are independent. This implies, \begin{align*} \Pr(A_i \cap A_j) = \Pr(A_i) \Pr(A_j) = p^6 \cdot p^6 = p^{12}. \end{align*} ($\Delta$ does not use this value, but I am just adding it for completeness).
$i$-th and $j$-th subsets have exactly one common node: There are $\binom{n}{4} \binom{4}{1} \binom{n-4}{3}$ subsets. In this case, you still want $12$ edges to exist in the graph that ensure $A_i$ and $A_j$ both hold. You can visualize this immediately by drawing two fully connected graphs, each with 4 nodes and one common node. Thus, $\Pr(A_i \cap A_j) = p^{12}$.
$i$-th and $j$-th subsets have two common nodes: There are $\binom{n}{4} \binom{4}{2} \binom{n-4}{2}$ subsets. In this case, note that since there are two common nodes, there is common edge. As a result, you only want $11$ edges to exist (as you don't count the common edge twice). This implies, $\Pr(A_i \cap A_j) = p^{11}$.
$i$-th and $j$-th subsets have three common nodes: There are $\binom{n}{4} \binom{4}{3} \binom{n-4}{1}$ subsets. Three common nodes imply three common edges and hence, $\Pr(A_i \cap A_j) = p^{9}$.
Thus, we can combine all this to obtain, \begin{align*} \Delta = \frac{1}{2}\binom{n}{4} \left( \binom{4}{1} \binom{n-4}{3} p^{12} + \binom{4}{2} \binom{n-4}{2} p^{11} + \binom{4}{3} \binom{n-4}{1} p^{9}\right). \end{align*} The $1/2$ is to account for the fact that $(i,j)$ and $(j,i)$ are counted once. On plugging in the value of $p$ and using some elementary algebra, we can show $\Delta = O(1/n)$. On the other hand, $\lim_{n \to \infty} \mu = \frac{c^{6}}{24}$.
Thus, Janson's inequality gives us \begin{align*} \lim_{n \to \infty} \Pr(X = 0) & \leq \lim_{n \to \infty} \exp (- \mu + O(1/n)) = \exp \left( -\frac{c^6}{24} \right). \end{align*}
Extending this to $k$ just requires one to consider multiple cases. It might be easier to find some pattern and show that $\Delta$ is $O(1/n)$ maybe using induction or something.