= Solution
Use the first <order statistic> and correct its bias:
$$
\widehat\theta(\mathbf X)=X_{(1)}-\frac1{n+1},\qquad X_{(1)}=\min_{1\leq i\leq n}X_i.
$$
This is a <measurable function> of the sample. Under $U_\theta^n$, write $Z=X_{(1)}-\theta$. It lies in $[0,1]$, and <independence> gives
$$
\mathbb P_\theta(Z>u)
=\prod_{i=1}^n\mathbb P_\theta(X_i>\theta+u)
=(1-u)^n,\qquad0\leq u\leq1.
$$
For a bounded nonnegative <random variable>, integrating the identities $Z=\int_0^1\mathbf1_{\{Z>u\}}\,du$ and $Z^2=\int_0^1 2u\mathbf1_{\{Z>u\}}\,du$ proves the <tail integral formula for moments> by <Tonelli's theorem>. Hence
$$
\begin{aligned}
\mathbb E_\theta Z&=\int_0^1(1-u)^n\,du=\frac1{n+1},\\
\mathbb E_\theta Z^2&=2\int_0^1u(1-u)^n\,du
=2\left(\frac1{n+1}-\frac1{n+2}\right)
=\frac2{(n+1)(n+2)}.
\end{aligned}
$$
Thus $\widehat\theta$ is an <unbiased estimator>. Its <quadratic risk> is the <variance> of $Z$, independent of $\theta$:
$$
\mathbb E_\theta(\widehat\theta-\theta)^2
=\frac2{(n+1)(n+2)}-\frac1{(n+1)^2}
=\frac{n}{(n+1)^2(n+2)}.
$$
Taking the <supremum> over all real $\theta$ and then the <infimum> over <estimators> gives
$$
\boxed{m_n\leq\frac{n}{(n+1)^2(n+2)}\leq\frac1{n^2},\qquad C'=1.}
$$
The final inequality follows from $n^3\leq(n+1)^2(n+2)$. Together with part (i), this proves that the <minimax risk> has order $n^{-2}$ for all $n\geq1$. In the last line of the source, the request for the upper-bound constant calls it $C$; here it is denoted $C'$ consistently with the upper-bound inequality.
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