= Solution
For $0<a<x<b$, put $S=\tau_a\wedge\tau_b$. The process $\phi(\|B_{t\wedge S}\|)$, where $\phi(r)=r^{2-d}$, is a bounded <martingale> by part (a). Also $S<\infty$ almost surely: before exit the endpoint $B_t$ lies inside the radius-$b$ ball, whose Gaussian probability tends to zero as $t\to\infty$. Path continuity makes the terminal radius either $a$ or $b$. If $p=\mathbb P_{\bar x}(\tau_a<\tau_b)$, the <dominated convergence theorem> applied to this bounded stopped <martingale> gives
$$
\phi(x)=p\phi(a)+(1-p)\phi(b).
$$
Solving yields
$$
\boxed{\mathbb P_{\bar x}(\tau_a<\tau_b)=\frac{\phi(b)-\phi(x)}{\phi(b)-\phi(a)}.}
$$
The denominator is nonzero because $\phi$ is strictly decreasing for $d\geq3$.
As $b\uparrow\infty$, the events $\{\tau_a<\tau_b\}$ increase to $\{\tau_a<\infty\}$. In one direction membership already implies a finite hit of $a$; in the other direction a continuous path up to its finite hit of $a$ has a finite maximum radius, so it lies in the event for every sufficiently large $b$. Continuity from below of probability and $\phi(b)\to0$ now give
$$
\boxed{\mathbb P_{\bar x}(\tau_a<\infty)=\frac{x^{2-d}}{a^{2-d}}=\left(\frac ax\right)^{d-2}.}
$$
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