= Solution
Set $\theta_0=1/2$ and $z=(\theta-1/2)^2$. Since
$$
\theta(1-\theta)=\frac14-z,
\qquad
0\leq z<\frac14,
$$
the risk is affine in $z$:
$$
g_{w,1/2}(\theta)
=\frac{w^2}{4n}
+z\left((1-w)^2-\frac{w^2}{n}\right).
$$
Its supremum is therefore controlled by the midpoint $z=0$ and the endpoint limit $z\to1/4$. Requiring both endpoint values to be at most the MLE's maximal risk $1/(4n)$ gives
$$
\frac{w^2}{4n}\leq\frac1{4n},
\qquad
\frac{(1-w)^2}{4}\leq\frac1{4n}.
$$
Equivalently,
$$
|w|\leq1,
\qquad
|1-w|\leq\frac1{\sqrt n}.
$$
Their intersection is
$$
\boxed{1-\frac1{\sqrt n}\leq w\leq1}.
$$
For precisely this range, the shrinkage estimator has maximal <mean squared error> no greater than that of $\widehat\theta$.
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