= Solution
This relative <absolute-error loss> weights each parameter by $1/\theta$. Its <posterior expected loss> is
$$
R_x(a)=b^2\int_0^\infty|\theta-a|e^{-b\theta}d\theta.
$$
The effective density for the <Bayes estimator under weighted absolute loss> is proportional to $\pi(\theta\mid x)/\theta$, hence an <exponential distribution> of rate $b$. Its median solves $1-e^{-ba}=1/2$.
More explicitly, for $a\geq0$, differentiating the split integrals at $\theta=a$ gives $R_x'(a)=b(1-2e^{-ba})$ and $R_x''(a)=2b^2e^{-ba}>0$. For $a<0$, the loss has derivative $-b$, so its minimum cannot lie there. The unique global minimum is therefore
$$
\boxed{\widehat\theta_B(X)=\frac{\log2}{\mu+X}.}
$$
This is the <weighted posterior median>; the ordinary median of the Gamma posterior would minimize unweighted absolute loss and would give a different answer.
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