= Solution
For $\delta(X)=X$ and $X\sim\operatorname{Pois}(\theta)$,
$$
R(\theta,\delta)
=\frac1\theta\mathbb E_\theta[(X-\theta)^2]
=\frac{\operatorname{Var}_\theta(X)}{\theta}
=1.
$$
Thus this rule has constant risk one.
To show that no rule has smaller worst-case risk, fix any $\alpha>1$ and let
$$
\pi_j=\operatorname{Gamma}(\alpha,\lambda_j),
\qquad \lambda_j\downarrow0.
$$
Part (a) gives
$$
\theta\mid X=x\sim
\operatorname{Gamma}(\alpha+x,\lambda_j+1).
$$
Since the reciprocal moment of a gamma variable with shape $k>1$ and rate $s$ is $s/(k-1)$, part (b)(iii) gives the Bayes rule
$$
\delta_j(x)=\frac{\alpha+x-1}{\lambda_j+1}.
$$
Its minimum posterior expected loss is
$$
\mathbb E(\theta\mid X=x)
-\big(\mathbb E(\theta^{-1}\mid X=x)\big)^{-1}
=\frac{\alpha+x}{\lambda_j+1}
-\frac{\alpha+x-1}{\lambda_j+1}
=\frac1{\lambda_j+1}.
$$
Hence the Bayes risk is
$$
r_j=\frac1{\lambda_j+1}\longrightarrow1.
$$
The criterion from part (b)(ii) now proves
$$
\boxed{\delta(X)=X\text{ is minimax}.}
$$
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