Need help in finding the asymptotic variance of an estimator. I kinda doing some review questions for my finals and I kinda got stuck on this question.

I'm able to do part a by finding the maximum likelihood estimator but for some reason.
To find the variance I used $Var(\theta)= E(\theta^2)-[E(\theta)]^2$
However given how the estimator of theta is written in b) I'm not sure how to find E$(\theta)$. 
 A: Hint: Find the information $I(\theta_0)$ for each estimator $\theta_0$. Then the asymptotic variance is defined as $$\frac{1}{nI(\theta_0 \mid n=1)}\,$$
for large enough $n$ (i.e., becomes more accurate as $n \to \infty$). Recall the definition of the Fisher information of an estimator $\theta$ given a density (probability law) $f$ for a random observation $X$:
$$I(\theta):=\mathbb{E}\left(\frac{\partial}{\partial \theta}\text{log}f(X \mid \theta)\right)^2 \,.$$
Under certain regularity conditions (which apply here), the information can be shown to be given by (this is much easier to compute than the above)
$$I(\theta)=-\mathbb{E}\left(\frac{\partial^2}{\partial \theta^2}\text{log}f(X \mid \theta)\right) \,.$$
Let us begin the computation for part (b). If you have computed the maximum likelihood estimator from part (a), then you should know that the log-likelihood function (for a single observation) is given by
$$\ell(\theta)=\log f(X \mid \theta) = \log\left(\frac{1}{\theta}\exp\left(-\frac{X}{\theta}\right)\right)=-\log(\theta)-\frac{X}{\theta} \,.$$
Now differentiating twice with respect to $\theta$, we obtain
$$\ell \,'(\theta)=-\frac{1}{\theta}+\frac{X}{\theta^{\,2}} \implies \ell \,''(\theta)=\frac{1}{\theta^{\,2}}-2\frac{X}{\theta^{\,3}}\,.$$
Now, as an aside, note that with $n=1$, we can write
$$\tilde{\theta} = \sqrt{\frac{X^2}{2}} = \frac{X}{\sqrt{2}}$$
where the second equality comes from the fact that $X \geq 0$.
Substituting $\theta=\tilde{\theta}$ (given $n=1$) and assuming $X \neq 0$, we have
$$\begin{align} \ell \,''(\tilde{\theta})&=\frac{2}{X^2}-2\frac{X}{\left(\frac{X}{\sqrt{2}}\right)^3} \\&= \frac{2}{X^2}-\frac{4\sqrt{2}}{X^2} \\&=2\frac{1-2\sqrt{2}}{X^2}\,. \end{align}$$
Thus, the information is given by
$$I(\tilde{\theta} \mid n=1) = -\mathbb{E}\left(2\frac{1-2\sqrt{2}}{X^2}\right)=2(2\sqrt{2} - 1)\cdot \mathbb{E}\left(X^{-2}\right) \,.$$
It remains to compute the expectation of $X^{-2}$ (allowing you to obtain the information and subsequently determine the asymptotic variance), which I will leave to you.
