Find UMVUE of $6\theta^2$ given $f(x\mid\theta) = \frac{1}{2\theta^2} e^{\frac{-\sqrt{x}}{\theta}} I_{(0,\infty)}(x)$ Given $X_1, X_2,\ldots, X_n$ are i.i.d  rvs with pdf $f(x\mid\theta) = \frac{1}{2\theta^2} e^{\frac{-\sqrt{x}}{\theta}} I_{(0,\infty)}(x)$  for $\ \theta > 0$.
(a) Find the UMVUE of $\ 6\theta^2$, and the variance of UMVUE.
(b) Find the CRLB of the UMVUE
(c) Find the variance of MME of $6\theta^2$ based on first moment.
My thought: For part (a), I got the same answer using Lehmann-Scheffer theorem as that of NCh's answer below. I would now present my computation for variance of UMVUE $= \frac{3}{2n^2+n} (\sum_{i=1}^{n} \sqrt{X_i})^2$. 
We have: $Var(\frac{3}{2n^2+n} (\sum_{i=1}^{n} \sqrt{X_i})^2) = \frac{9}{(2n^2+n)^2} (E(Z^4) - E^2(Z^2))$  where $Z = \sum_{i=1}^{n} \sqrt{X_i}$. Now, $E(Z^2) = \theta^2(4n^2+2n)$. In addition, $Z$ is the sum of $n$ i.i.d r.vs which follow $N(\ 2n\theta, \ 2n\theta^2)$, so $E(Z^4) = (2n\theta)^4 + 6(2n\theta)^2(2n\theta^2)^2 + 3(2n\theta^2)^4$. 
Thus, $\frac{9}{(2n^2+n)^2} (E(Z^4) - E^2(Z^2)) = \frac{9}{(2n^2+n)^2} (16n^4\theta^4 + 96n^4\theta^6 + 48\theta^8n^4 - 16n^4\theta^4 - 16n^3\theta^4 - 4n^2\theta^4)$
(b) To find the CRLB, I need first to verify $\ E\ (\overline{X}\frac{\partial}{\partial\theta}\log f(\overline{X}\mid\theta)) = 12\theta,$ as $\overline{X}$ is an unbiased estimator of $6\theta$. Now, $\log f(\overline{X}\mid\theta) = \frac{-\sqrt{\overline{X}}}{\theta} - \log{2\theta^2}$, so $E\ (\overline{X}\frac{\partial}{\partial\theta}\log f(\overline{X}\mid\theta)) = E(\frac{\overline{X}^{3/2} - 2\theta\overline{X}}{\theta^2}) = -12\theta + \frac{1}{\theta^2}E(\overline{X}^{3/2}).$ But I got stuck here since I cannot compute $\ E(\overline{X}^{3/2})$.
(c) The MME of $6\theta^2$ based on first moment is $\overline{X}$. Thus, we only need to compute $Var(\overline{X}) = \frac{Var(X_1)}{n} = \fbox{$\frac{84\theta^4}{n}$}$ (since $E(X^2) = 120\theta^4$)
My question: Could someone please help with parts (a) and (b) above? Also, is my answer for part (c) correct?

It is no need to compute conditional expectations. Let $$T=\overline{\sqrt{X}}=\frac{\sum_{i=1}^n \sqrt{X_i}}{n}$$
You have already found its expectation:
$$\mathbb ET=\mathbb E\sqrt{X_1}=2\theta.$$
Let us find UMVUE of $6\theta^2$ as $\theta^*=c(n)T^2$. Firstly find $\mathbb E T^2$:
$$
\mathbb ET^2 = \text{Var}\,T+\left(\mathbb ET\right)^2 = \frac{\text{Var}\sqrt{X_1}}{n}+4\theta^2=\frac{\mathbb EX_1-4\theta^2}{n}+4\theta^2 = \frac{2\theta^2}{n}+4\theta^2.
$$ 
And if we take $$
c(n) = \frac{6}{\frac2n+4}=\frac{3n}{1+2n}$$
we'll obtain the desired result: 
$$
\mathbb E\theta^* = \mathbb E\left(\frac{6}{\frac2n+4}T^2\right) = \frac{6}{\frac2n+4} \times \left(\frac{2\theta^2}{n}+4\theta^2\right)=6\theta^2.
$$
And $\theta^*$ is UMVUE since it is a function of complete sufficient statistics. 
