We are in $\mathbb R^2$. Let $A_k=\{x\mid R<|x|<2^k R\}$ and $T_K=\inf \{t>0\mid |B_t|<R\}$ where $(B_t)$ is a Brownian in $\mathbb R^2$ starting at $2^kR>|b|>R$. Let $f\in \mathcal C^2(\mathbb R)$ with compact support s.t. $f(x)=-\log(x)$ whenever $x\in A_k$. Let $\alpha _k$ the first time of exist of $A_k$. By Dynkin formula, $$\mathbb E^b[f(B_{\alpha _k})]=f(b).\tag{e}$$ I set $p_k=\mathbb P^b(|B_{\alpha _k}|=R)$ and $q_k=\mathbb P(|B_{\alpha _k}|=2^kR)$. By $(e)$, $$-\log(R)\cdot p_k-(\log(R)+k\log(2))q_k=-\log|b|,\tag{*}$$ for all $k$. Therefore, $q_k\to 0$ when $k\to \infty $ and thus $\mathbb P(T_K<\infty )=1$.
Question : Why $(*)$ give us $q_k\to 0$ when $k\to \infty $ and why this implies that $\mathbb P(T_K<\infty )=1$ ?
You have that $p_k=1-q_k$ and thus $(*)$ gives you $$-\log(R)+k\log(2)q_k=-\log |b|.$$ So if $q_k\not\to 0$, then the LHS $\to \infty $ which is not possible.
Because $\{T_K=\infty \}=\bigcap_{k\in\mathbb N}\{|B_{\alpha _k}|=2^kR\},$ and thus $$\mathbb P\{T_K=\infty \}=\lim_{k\to \infty }q_k=0.$$