= Solution
Write $u(z)=|z|^{2-d}$. For a radial function $r^{2-d}$, the <Laplacian> is
$$
u''(r)+\frac{d-1}{r}u'(r)=(2-d)(1-d)r^{-d}+(d-1)(2-d)r^{-d}=0.
$$
Thus $u$ is a <harmonic function> off zero. Fix $R>|x|$ and stop <Brownian motion> when it reaches either sphere of radius $\varepsilon$ or $R$, at time $S_R$. This exit time is finite: stopping the <martingale> $|B_t|^2-dt$ at $S_R\wedge t$ yields $d\mathbb E(S_R\wedge t)\leq R^2-|x|^2$, and <monotone convergence> applies. By <Itô's formula>, $u(B_{t\wedge S_R})$ is a bounded <martingale>. <Optional stopping> and <dominated convergence> therefore give
$$
u(x)=q_R\varepsilon^{2-d}+(1-q_R)R^{2-d},\qquad
q_R=\mathbb P_x(T_\varepsilon<T_R)=\frac{|x|^{2-d}-R^{2-d}}{\varepsilon^{2-d}-R^{2-d}}.
$$
The events $\{T_\varepsilon<T_R\}$ increase to $\{T_\varepsilon<\infty\}$ as $R\to\infty$: a continuous path up to any finite hitting time is bounded. Since $d\geq3$, $R^{2-d}\to0$, proving the <Brownian hitting probability of a ball>
$$
\boxed{\mathbb P_x(T_\varepsilon<\infty)=(\varepsilon/|x|)^{d-2}.}
$$
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