= Solution
Let $\tau=S_r\wedge T_R$. The stopped logarithm lies between $\log r$ and $\log R$, so the <bounded local martingale criterion> makes it a true bounded <martingale>. Also $\tau$ is finite almost surely. Indeed $|B_t|^2-2t$ is a <martingale>, and applying its stopped identity through $t\wedge T_R$ gives
$$
2\mathbb E(t\wedge T_R)=\mathbb E|B_{t\wedge T_R}|^2-1\leq R^2-1.
$$
<monotone convergence theorem> implies $\mathbb E T_R\leq(R^2-1)/2<\infty$, hence $\tau\leq T_R<\infty$ almost surely. Boundedness permits <dominated convergence theorem> in the stopped logarithm as $t\to\infty$. Consequently
$$
\boxed{\mathbb E\log|B_{S_r\wedge T_R}|=0.}
$$
The standard results used are Itô's formula, the bounded-local-martingale criterion, the stopped <martingale> identity at bounded <stopping times>, and dominated/<monotone convergence theorem>. At exit the radius is either $r$ or $R$ by continuity. Solving $p\log r+(1-p)\log R=0$ also gives the <planar Brownian annulus hitting probability>
$$
\mathbb P(S_r\leq T_R)=\frac{\log R}{\log R-\log r}.
$$
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