= Solution
{c}
For the new integrand, $\theta=\mathbb E_p[e^{-X_1}]=e^{1/2}$ and $\operatorname{Var}_p(e^{-X_1})=e^2-e$, independently of dimension. <Chebyshev's inequality> therefore yields
$$
\mathbb P\!\left(|\widehat\theta_n^{\mathrm{RS}}-\theta|\geq\varepsilon\right)
\leq\frac{e^2-e}{n\varepsilon^2}.
$$
A sufficient positive integer sample count is
$$
\boxed{n=\left\lceil\frac{e^2-e}{\delta\varepsilon^2}\right\rceil.}
$$
The <ceiling function> ensures that the bound is at most $\delta$, so the strict-error event in the question has <probability> at least $1-\delta$. The required number of accepted draws is independent of $d$; their expected proposal count is $n\sigma^d$.
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