= Solution
Use two translated <uniform distributions>, with parameters $0$ and $a=1/(2n)$. Their distance in the usual <metric> is $a$, so take the separation radius $\delta=a/2=1/(4n)$ and the <loss function> $\Phi(r)=r^2$. The supplied testing reduction, equivalently the <Le Cam two-point lemma>, and part (a)(i) imply
$$
m_n\geq\delta^2\frac{1-V(U_0^n,U_a^n)}2
=\frac{a^2}{8}\bigl[1-V(U_0^n,U_a^n)\bigr].
$$
To see the reduction directly, turn any <estimator> into a binary <hypothesis test> by choosing the closer of $0$ and $a$. If the test is wrong, the absolute estimation error is at least $a/2$. The worst <quadratic risk> is at least its average over these two parameters, giving precisely this bound.
Under <Lebesgue measure> on $\mathbb R^n$, the two sample <probability density functions> are the indicators of the cubes $[0,1]^n$ and $[a,a+1]^n$. Their intersection is $[a,1]^n$, of volume $(1-a)^n$. Therefore
$$
1-V(U_0^n,U_a^n)
=\int\min(p_0^{(n)},p_a^{(n)})\,d\mathbf x
=(1-a)^n\geq1-na=\frac12,
$$
where the inequality is <Bernoulli's inequality>. Consequently
$$
\boxed{m_n\geq\frac{a^2}{16}=\frac1{64n^2},\qquad C=\frac1{64}.}
$$
This also fits the <Hellinger distance> approach: the two one-sample affinities give $H^2(U_0,U_a)=a$, while equal-height indicator densities imply $V(U_0^n,U_a^n)=H^2(U_0^n,U_a^n)$. Part (a)(ii) thus yields the same bound $V\leq na=1/2$. The change of support produces <Squared Hellinger distance> proportional to $a$, explaining the parameter scale $a\asymp n^{-1}$ and <quadratic risk> scale $n^{-2}$.
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