= Solution
Write $I_1(\theta)$ for the <Fisher information> in one observation:
$$
I_1(\theta)=\mathbb E_\theta\!\left[\left\{\frac{\partial}{\partial\theta}\log f(X;\theta)\right\}^2\right]
=-\mathbb E_\theta\!\left[\frac{\partial^2}{\partial\theta^2}\log f(X;\theta)\right].
$$
Under the stipulated regularity assumptions, the <maximum-likelihood estimator> is <consistent> and has <asymptotic normality>:
$$
\boxed{\sqrt n(\widehat\theta-\theta)\ \xrightarrow{d}\ N\!\left(0,I_1(\theta)^{-1}\right).}
$$
Equivalently, its large-sample <normal approximation> has <mean> $\theta$ and <variance> $1/[nI_1(\theta)]$. The information for the whole <random sample> is $nI_1(\theta)$; keeping the one-observation convention explicit prevents an extra factor of $n$.
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