= Solution
The denominator $\Phi(z)$ is positive, so
$$
B<\frac1{10}
\quad\Longleftrightarrow\quad
1-\Phi(z)<\frac1{10}\Phi(z)
\quad\Longleftrightarrow\quad
\Phi(z)>\frac{10}{11}.
$$
Strict increase of the <standard normal distribution> function turns this into $z>\Phi^{-1}(10/11)$. Since the coefficient multiplying $\overline x$ in $z$ is positive, the rejection rule is precisely $\overline x>c$, with
$$
\boxed{c=\frac{\sigma\sqrt{\sigma^2+n\tau^2}}{n\tau}
\Phi^{-1}\!\left(\frac{10}{11}\right)
=\frac\sigma{\sqrt n}\sqrt{1+\frac{\sigma^2}{n\tau^2}}
\Phi^{-1}\!\left(\frac{10}{11}\right).}
$$
The <quantile> is approximately $1.33518$. Equivalently, the rule requires <posterior probability> of $H_0$ below $1/11$, because its <prior odds> are one. The calculation is an instance of the <Bayes factor for the sign of a normal mean>.
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