 Solution

The MLE is $\hat{\mu }=\bar{X}$ with MSE $1/n$. Combining likelihood and the $N(0,{\tau }^{-2})$ prior gives posterior $N(n\bar{X}/(n+{\tau }^2),1/(n+{\tau }^2))$. The frequentist MSE of its mean is $[n+{\tau }^4{\mu }^2]/(n+{\tau }^2{)}^2$, which is below $1/n$ exactly when ${\mu }^2<2{\tau }^{-2}+1/n$.