= Solution
The Neyman–Pearson lemma says that among tests of size at most $\alpha$ for two simple hypotheses, a <likelihood-ratio test> rejecting where $p(x;\theta_1)/p(x;\theta_0)>k$ is most powerful, with boundary randomisation if required. If $\varphi$ is that test and $\psi$ any competing test, choose $k$ so the sizes agree. Pointwise,
$$
(\varphi-\psi)(p_1-kp_0)\ge0.
$$
Integration and the size inequality yield $E_1\varphi\ge E_1\psi$.
Here
$$
\frac{p(x;\theta_1)}{p(x;\theta_0)}=\frac{\theta_1}{\theta_0}e^{-(\theta_1-\theta_0)|x|},
$$
which decreases with $|x|$. Thus reject for $|X|\le c$, where
$$
\boxed{\alpha=P_{\theta_0}(|X|\le c)=1-e^{-\theta_0c},\qquad
c=-\frac1{\theta_0}\log(1-\alpha).}
$$
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