= Solution
The <Rao-Blackwell theorem> says that, for a <sufficient statistic> $T$ and a finite-second-moment estimator $\delta(X)$, the conditional estimator $\delta_T(T)=E_\theta[\delta(X)\mid T]$ can be chosen independently of the unknown parameter and has no greater <mean squared error> for estimating any specified function $g(\theta)$. Sufficiency makes the <conditional distribution> of the data given $T$ parameter-independent. <Conditional expectation> preserves the <expectation>, hence preserves unbiasedness. Writing $\delta-g=(\delta-\delta_T)+(\delta_T-g)$, the cross term has <expectation> zero, because $E[\delta-\delta_T\mid T]=0$. Therefore
$$
E_\theta[(\delta-g)^2]
=E_\theta[(\delta_T-g)^2]+E_\theta[(\delta-\delta_T)^2]
\ge E_\theta[(\delta_T-g)^2].
$$
Equality holds exactly when $\delta=\delta_T$ almost surely. This proves the squared-loss form, including the usual unbiased-estimator <variance> statement.
Here $\theta>0$ is necessary for the stated uniform interval. Put $L=X_{(1)}=\min_iX_i$ and $R=X_{(n)}=\max_iX_i$. The joint density factors as
$$
p_\theta(x)=(2\theta)^{-n}\mathbf1\{\theta<L,\ R<3\theta\}
=(2\theta)^{-n}\mathbf1\{R/3<\theta<L\}.
$$
The <factorization theorem for sufficient statistics> shows that \b[$T=(L,R)$ is sufficient]. Also $E[X_1]=2\theta$, so $X_1/2$ is unbiased and has <variance> $\theta^2/12$.
For $n\ge2$, conditional on distinct extrema $l,r$, <exchangeability> gives probability $1/n$ that $X_1$ is each endpoint; with probability $(n-2)/n$ it is one of the remaining observations, uniform on $(l,r)$. Its conditional mean is consequently
$$
E[X_1\mid L=l,R=r]=\frac{l+r}{n}+\frac{n-2}{n}\frac{l+r}{2}=\frac{l+r}{2}.
$$
Thus the <Rao-Blackwell estimator for a uniform scale interval> is
$$
\boxed{\widetilde\theta=\frac{L+R}{4}},
$$
which is unbiased and has no greater <mean squared error>. For $n=1$, $L=R=X_1$ and the same formula reduces to the original estimator. As a quantitative check, the <covariance of two uniform order statistics> gives
$$
\operatorname{Var}(\widetilde\theta)=\frac{\theta^2}{2(n+1)(n+2)}\le\frac{\theta^2}{12}.
$$
For instance, after writing $X_i=\theta+2\theta U_i$, both extreme uniform <variances> are $n/[(n+1)^2(n+2)]$ and their <covariance> is $1/[(n+1)^2(n+2)]$. These formulas follow by integrating the joint extreme density $n(n-1)(r-l)^{n-2}$ on $0<l<r<1$; they also display strict improvement for $n>1$.
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