= Solution
Put $S=\sum_iX_i$, and let $z_{.95}$ be the exact standard-normal 95th percentile. Under parameter $\theta$, $S=n\theta+\sqrt n Z$ with $Z$ standard normal. This distribution can be justified directly: the joint density of the centered observations is proportional to $\exp[-\sum_i(X_i-\theta)^2/2]$. An orthogonal change of coordinates with first row $(1,\ldots,1)/\sqrt n$ has unit absolute Jacobian and preserves the sum of squares. Its density factors into $n$ one-dimensional standard-normal densities, making the first coordinate $Z=(S-n\theta)/\sqrt n$ standard normal.
Consider the test $\phi_*=\boldsymbol1_{\{S>\sqrt n\,z_{.95}\}}$. Its rejection probability is
$$
\Pr_\theta(\phi_*=1)=1-\Phi(z_{.95}-\sqrt n\theta),
$$
increasing in $\theta$. Therefore its maximum rejection probability over the null $\theta\leq0$ is achieved at zero and equals $\alpha=1/20$. This proves it has the required <test size>.
To prove the <uniformly most powerful test> property, rather than quote the <Neyman-Pearson lemma>, fix any $\theta_1>0$. The <likelihood ratio> of the full sample against $\theta=0$ is
$$
\frac{p_{\theta_1}(x)}{p_0(x)}=\exp\left(\theta_1S-\frac{n\theta_1^2}{2}\right),
$$
strictly increasing in $S$. Set $K=\exp(\theta_1\sqrt n z_{.95}-n\theta_1^2/2)$. For any competing possibly randomized test $0\leq\phi\leq1$ with composite-null size at most $\alpha$, we have $\mathbb E_0\phi\leq\alpha=\mathbb E_0\phi_*$. Pointwise,
$$
(\phi_* -\phi)(p_{\theta_1}-Kp_0)\geq0,
$$
because both factors are nonnegative above the threshold and both are nonpositive below it. Integrating gives
$$
\mathbb E_{\theta_1}\phi_*-\mathbb E_{\theta_1}\phi\geq K(\mathbb E_0\phi_*-\mathbb E_0\phi)\geq0.
$$
This comparison proves the likelihood-ratio optimality result needed here in full. Since the same <critical region> works for every $\theta_1>0$, it proves the <uniformly most powerful test> property over the entire alternative.
For $n=9$, the exact <critical region> and its quoted-table approximation are
$$
\boxed{\sum_{i=1}^9X_i>3z_{.95},\quad\text{equivalently }\overline X>z_{.95}/3\approx0.55.}
$$
Using the supplied rounded percentile $1.65$ gives $\sum X_i>4.95$. The exact cutoff uses $z_{.95}\approx1.644854$ and hence $\overline X>0.548285$; the rounded cutoff is the intended numerical approximation, not a literally exact size-$0.05$ threshold. The Gaussian distribution is continuous, so equality at the cutoff has probability zero and no randomization there is needed.
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