I'm asked to calculate the mean square error of $\phi(x-\bar X_n)$ given a sample $X_1,\ldots,X_n$ from $\mathcal N(\mu,1)$. Here $\phi$ denotes the standard normal density. Now by the Delta method, using $\sqrt n(\bar X_n-\mu)\stackrel{\mathcal D}\to\mathcal N(0,1)$ and $g(x)=\phi(t-x)$, we find $$\sqrt n(\phi(t-\bar X_n)-\phi(t-\mu))\stackrel{\mathcal D}\to\mathcal N(0,(t-\mu)^2\phi^2(t-\mu)),$$ which implies that $MSE(\phi(t-\bar X_n))\to\frac1n(t-\mu)^2\phi^2(t-\mu)$, as $n\to\infty$.
However, I'm asked for the mean square error of $\phi(x-\bar X_n)$, not for its asymptotic value. It seems hard to answer this question. Is it likely that I'm asked for asymptotic value, but that this condition is forgotten in the exercise statement? Or is it possible to calculate this MSE, for arbitrary $n\in\mathbb N$?