= Solution
Let
$$
a=(-1,2,0,4),\qquad\|a\|_2^2=21,
$$
so $T(\theta)=a^T(\theta_1,\ldots,\theta_4)$. In the <Gaussian sequence model>, put $h_k=\sqrt n(\theta_k-Y_k)$. For each relevant coordinate, the posterior density of $h_k$ relative to the standard-normal density $\varphi$ is
$$
\frac{\pi(Y_k+h_k/\sqrt n)}
{\int\pi(Y_k+u/\sqrt n)\varphi(u)du}.
$$
The log-Lipschitz assumption implies
$$
e^{-c|h_k|/\sqrt n}
\leq\frac{\pi(Y_k+h_k/\sqrt n)}{\pi(Y_k)}
\leq e^{c|h_k|/\sqrt n}.
$$
These bounds provide Gaussian-integrable domination, while the ratio converges pointwise to one. Dominated convergence, coordinate independence, and the same argument after multiplying by $e^{ta^Th}$ show that, under the posterior,
$$
\mathbb E^\Pi\left[
e^{t\sqrt n\{T(\theta)-T(Y)\}}\mid Y
\right]
\longrightarrow e^{21t^2/2}
$$
almost surely. The supplied moment-generating-function criterion therefore gives the finite-functional <Bernstein-von Mises theorem>
$$
\sqrt n\{T(\theta)-T(Y)\}\mid Y
\Longrightarrow N(0,21),
$$
with uniform convergence of distribution functions.
If $z_{1-\alpha}=\Phi^{-1}(1-\alpha)$, the posterior quantile defining $R_n$ consequently satisfies
$$
\boxed{\sqrt nR_n\longrightarrow
\sqrt{21}\,z_{1-\alpha}}
$$
in probability. Under $P_{\theta_0}^Y$,
$$
\sqrt n\{T(\theta_0)-T(Y)\}=-a^Tg\sim N(0,21)
$$
for every $n$. Quantile convergence and the <Slutsky theorem> now yield
$$
\mathbb P_{\theta_0}^Y(T(\theta_0)\in C_n)
\longrightarrow1-\alpha.
$$
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