= Solution
Interpret the printed expression as $d^*(x)=x/(\sigma^2\lambda)$. If $X\sim N(\theta,\sigma^2)$, the Gaussian exponential-moment formula gives
$$
R(\theta,d^*)=\mathbb E_\theta e^{-\theta X/\sigma^2}=\exp\left(-\frac{\theta^2}{\sigma^2}+\frac{\theta^2}{2\sigma^2}\right)=e^{-\theta^2/(2\sigma^2)}.
$$
Its supremum is one, attained at zero. For every rule $d$, $L(0,d(X))=1$, so $R(0,d)=1$ and $\sup_\theta R(\theta,d)\geq1$. Hence
$$
\boxed{d^*(x)=\frac{x}{\sigma^2\lambda}\text{ is minimax, with minimax risk }1.}
$$
In the language of part (b), the point-mass prior $\delta_0$ makes every rule Bayes with <Bayes risk> one, and $d^*$ has exactly that worst-case risk.
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