Solution 1:

A complete sufficient statistic for $\mu$ is $\overline X$, or equivalently $S=\sum\limits_{i=1}^n X_i$.

Now $(X_1,S)$ has a bivariate normal distribution (can be verified quickly using moment generating functions), more specifically $(X_1,S)\sim N_2\left(\mu\begin{pmatrix}1\\n\end{pmatrix},9\begin{pmatrix}1&1\\1& n\end{pmatrix}\right)$.

This gives the conditional distribution of $X_1$ given $S=s$ as a univariate normal $N\left(\frac sn,9\left(1-\frac1n\right)\right)$.

Hence the UMVUE of $P(X_1\le c)$ is

\begin{align} E_{\mu}\left[I_{X_1\le c}\mid S\right]&=P_{\mu}(X_1\le c\mid S) \\&=\Phi\left(\frac{c-S/n}{3\sqrt{1-1/n}}\right) \\&=\Phi\left(\frac{c-\overline X}{3\sqrt{1-1/n}}\right) \end{align}