It is not difficult to show, using standard concentration bounds, that for $G\sim G(n,p)$ with $p=f(n)\cdot\ln{n}/n$, $f(n)\ge 1+\varepsilon$, $\varepsilon>0$ constant, that $G$ is nearly regular. More formally, \begin{equation*} \delta(G),\Delta(G)=\Theta(np). \end{equation*}
In fact, it is not difficult to strengthen that statements so it would contain the specific constants $c_1\le 1\le c_2$ such that for every $v\in V(G)$, $c_1np \le d(v) \le c_2np$. If we also know that $f(n)=\omega(1)$, then $\delta(G),\Delta(G)=np(1+o(1))$.
In a book by Bollobás, there's a strong statements according to which if $p=p(n)\le (1+o(1))\ln{n}/n$, then with high probability $\delta(G)=d(n)$ or $\delta(G)=d(n)+1$ for some predetermined $d(n)$, but it is not mentioned what is $d(n)$.
My attempt is as follows: I let $X_k$ count the number of vertices of degree $k$. I show that for every $\delta>0$, $\mathbb{E}(X_{\delta\ln{n}})\to\infty$. That's not enough; in order to conclude the argument, I must also prove that $\text{Var}(X) = o(\mathbb{E}^2(X))$, and then I can finish with Chebyshev's inequality. However, as for every pair of distinct vertices $u$, $v$ the events "$d(u)=k$" and "$d(v)=k$" are dependent, the variance of $X_k$ is quite difficult to estimate.
Is that the way to go, or can you think of a better method in this case?