= Solution
The likelihood contributes precision $n$ and precision-weighted mean $n\overline X$, while the prior contributes precision $\tau^2$ and precision-weighted mean $\tau^2\theta$. By <normal-normal conjugacy with known observation variance>, the posterior is
$$
\mu\mid X_1,\ldots,X_n
\sim N\left(
\frac{n\overline X+\tau^2\theta}{n+\tau^2},
\frac1{n+\tau^2}\right).
$$
Thus the posterior mean is
$$
\boxed{\widehat\mu=\frac{n\overline X+\tau^2\theta}{n+\tau^2}}.
$$
When $\theta=0$, its expectation under true parameter $\mu$ is $n\mu/(n+\tau^2)$ and its variance is $n/(n+\tau^2)^2$. Hence its frequentist <mean squared error> is
$$
\boxed{\operatorname{MSE}_\mu(\widehat\mu)
=\frac{n+\tau^4\mu^2}{(n+\tau^2)^2}}.
$$
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