= Solution
Put $\kappa_i=n/(n+q_i)$. The <posterior mean> of the <Gaussian practical-null mixture> is
$$
\mathbb E[\beta\mid y]=[w_0(y)\kappa_0+w_1(y)\kappa_1]y.
$$
Near zero, $B_{01}(0)>1$, so the narrow component has high <posterior probability> and $\kappa_0$ is small when $n_0\gg n$. This pulls the <Bayesian model averaging> posterior toward zero. For the given ratios, writing $z=\sqrt n\,y$ gives
$$
B_{01}(y)=10e^{-99z^2/202},\qquad B_{01}>1\ \Longleftrightarrow\ |z|<2.16753\ldots.
$$
Thus the order statement $y=O(n^{-1/2})$ alone does not guarantee strong pull toward zero: $z=3$ already has the opposite model preference.
As $|y|$ grows, $V_1>V_0$ makes $B_{01}(y)\to0$, and the <mixture model> approaches $N(\kappa_1 y,1/(n+q_1))$. \b[Its center is exactly $\kappa_1 y$, and approximately $y$ only for a sufficiently diffuse wide prior], meaning $q_1\ll n$. Here $\kappa_1=100/101$, so the relative displacement is about one percent. Large observations alone do not remove this finite-prior shrinkage.
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