= Solution
Let $\rho=|z_0|$ and $f(x,y)=\log\sqrt{x^2+y^2}$. On the <annulus> $r<\sqrt{x^2+y^2}<R$, its <gradient> and <Laplace operator> are
$$
f_x=\frac{x}{x^2+y^2},\qquad f_y=\frac{y}{x^2+y^2},
\qquad \Delta f=0.
$$
Thus $f$ is a <harmonic function> there. The stopped <Itô formula> gives
$$
\log|B_{t\wedge\tau}|=\log\rho
+\int_0^{t\wedge\tau}\frac{B_1(s)}{|B_s|^2}\,dB_1(s)
+\int_0^{t\wedge\tau}\frac{B_2(s)}{|B_s|^2}\,dB_2(s).
$$
The sum of the squared coefficients is $|B_s|^{-2}\leq r^{-2}$ before $\tau$. The <Itô isometry> therefore makes the stochastic term a true <square-integrable> <martingale> on every finite interval. Continuity prevents overshoot at exit, so $\log r\leq\log|B_{t\wedge\tau}|\leq\log R$. This proves the required bounded <martingale> property.
Before taking the terminal <expectation>, prove that $\tau$ is finite. Set $\theta_R=\inf\{t\geq0:|B_t|=R\}$. The <Itô formula> for $|B|^2$ shows that
$$
|B_{t\wedge\theta_R}|^2-\rho^2-2(t\wedge\theta_R)
$$
is a <martingale>, since its stopped stochastic integrands are bounded. Taking <expectations> yields
$$
2\mathbb E(t\wedge\theta_R)
=\mathbb E|B_{t\wedge\theta_R}|^2-\rho^2\leq R^2-\rho^2.
$$
The <monotone convergence theorem> gives $\mathbb E\theta_R\leq(R^2-\rho^2)/2<\infty$. Hence $\tau\leq\theta_R<\infty$ <almost surely>.
The <bounded convergence theorem>, equivalently the <optional stopping theorem> applied to this bounded <martingale>, gives $\mathbb E\log|B_\tau|=\log\rho$. The two exit events partition the sample space up to a null set. Writing $p=\mathbb P(\tau_r<\tau_R)$,
$$
\log\rho=p\log r+(1-p)\log R.
$$
Solving gives the <planar Brownian annulus hitting probability>
$$
\boxed{\mathbb P(|B_\tau|=r)=\mathbb P(\tau_r<\tau_R)
=\frac{\log(R/\rho)}{\log(R/r)}.}
$$
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