To close this one:
The density is recognized as a Gamma distribution with shape parameter $k=3$ and unknown scale parameter $\theta$, so we have $E(X) = 3\theta$ and $\operatorname {Var}(X) = 3\theta^2$
Given an i.i.d sample, the expected value and variance of the MLE is then
$$E(\hat \theta_{MLE}) = \theta,\;\;\operatorname {Var}(\hat \theta_{MLE}) = \frac 1 {3n}\theta^2$$
For consistency, we can use then the sufficient conditions: $\lim_{n\rightarrow}E(\hat \theta_{MLE})=\theta$, holds, and $\lim_{n\rightarrow}\operatorname {Var}(\hat \theta_{MLE})=0 \Rightarrow \lim_{n\rightarrow}\frac 1 {3n}\theta^2=0$, holds too, so the MLE is consistent.
The Central Limit Theorem Holds and so
$$\sqrt n(\hat \theta_{MLE}-\theta) \rightarrow_d \mathcal N(0, \frac 13\theta^2)$$
The Fisher Information is
$$\mathcal I(\theta) = -E\left[\frac 3 {\theta^2}-2\frac X {\theta^3}\right] = -\frac 3 {\theta^2} + 2\frac {E(X)}{\theta^3} = -\frac 3 {\theta^2} + 2\frac {3\theta}{\theta^3} = \frac 3{\theta^2}$$
and so the MLE achieves the Cramer-Rao bound.
As for the question about minimum variance estimator, @cardinal's answer here
https://math.stackexchange.com/questions/28779/minimum-variance-unbiased-estimator-for-scale-parameter-of-a-certain-gamma-distr
is a complete and sufficient statistic for estimating the answer.
Best Answer
Yes, there is and it can be derived routinely. The Fisher Information can be shown to be
$$I(\rho) = \frac{1+\rho^2}{\left(1-\rho^2\right)^2}$$
and you know how to get the CRLB from here. The result may be arrived at simply by applying the definition of Fisher Information, i.e. start from the log likelihood
$$\log\left[ f(x;\rho) \right] =\log\left\{ \frac{1}{2\pi \sqrt{1-\rho^2}}\exp\left\{-\frac{1}{2(1-\rho^2)} \left(x^2 + y^2 - 2\rho xy \right) \right\} \right\}$$
take the derivatives and evaluate the expectation using the properties of the normal distribution. I advise you to verify it on your own.