= Solution
The <James–Stein estimator> in the specified unit-variance model is
$$
\boxed{\hat\theta_{\rm STEIN}(X)=\left(1-\frac{p-2}{\|X\|^2}\right)X},
$$
with an arbitrary value at $X=0$, a probability-zero event. The risk of $X$ is $p$. <Gaussian Stein identity> states $E_\theta[(X_i-\theta_i)g_i(X)]=E_\theta[\partial_i g_i(X)]$ for suitably integrable differentiable functions; it extends to the present singular function by truncation, since $p>2$ makes $E\|X\|^{-2}$ finite.
Put $c=p-2$ and $g(x)=-cx/\|x\|^2$. Then $\|g\|^2=c^2/\|x\|^2$ and $\operatorname{div}g=-c(p-2)/\|x\|^2$. Expanding the <quadratic risk> and applying Stein's lemma yields
$$
\boxed{R(\theta,\hat\theta)=p-(p-2)^2E_\theta\frac1{\|X\|^2}<p=R(\theta,X)}.
$$
Thus domination is strict for every finite $\theta$.
To identify the worst-case risk, let $r=\|\theta\|\to\infty$. On $\|X\|>r/2$, the reciprocal square is at most $4/r^2$. On the complementary ball, the Gaussian density is at most $(2\pi)^{-p/2}e^{-r^2/8}$, so its reciprocal-square integral is bounded by a constant times $e^{-r^2/8}r^{p-2}$, tending to zero. Therefore $E_\theta\|X\|^{-2}\to0$, and $R(\theta,\hat\theta)\to p$ from below. Hence
$$
\boxed{\sup_\theta R(\theta,X)=\sup_\theta R(\theta,\hat\theta)=p}.
$$
The second supremum is approached at infinity and need not be attained at a finite parameter.
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