= Solution
On a compact time interval a continuous path avoiding the origin has a strictly positive minimum radius. Thus the times $S_{1/n}$ tend to infinity almost surely. Part (a) localizes $M_t=\log|B_t|$, so $M$ is a <continuous local martingale>.
To show that it is not a <martingale>, compute its <expectation> at a positive deterministic time. In distribution $B_t=(1,0)+W_t$, where $W_t$ is a centered isotropic Gaussian vector. Conditional on $\rho=|W_t|$, its angle is uniform. The <angular average of a logarithmic potential> is
$$
\frac1{2\pi}\int_0^{2\pi}\log|1+\rho e^{i\theta}|\,d\theta=\log\max(1,\rho).
$$
For $\rho<1$ this follows by averaging the real part of the power series for $\log(1+\rho e^{i\theta})$; for $\rho>1$ factor out $\rho$ and apply the same argument to $\rho^{-1}$. The value at $1$ follows by the integrable logarithmic singularity, or is irrelevant for the continuously distributed radius.
The radius has density $(\rho/t)e^{-\rho^2/(2t)}$. The logarithm of $|B_t|$ is absolutely integrable: near the origin its Gaussian density is bounded and $\int_0^1r|\log r|\,dr<\infty$; away from the origin its positive part is bounded by a constant plus $|B_t|$. Therefore conditioning is legitimate, and integration by parts yields
$$
\boxed{\mathbb EM_t=\int_1^\infty\log\rho\,\frac\rho t e^{-\rho^2/(2t)}\,d\rho=\int_1^\infty\frac{e^{-\rho^2/(2t)}}\rho\,d\rho>0\quad(t>0).}
$$
Since $M_0=0$, this contradicts preservation of <expectation> by a <martingale>. Thus the <logarithmic radius of planar Brownian motion> is a strict local <martingale>. It is not a nonnegative local <martingale>; its negative values allow its <expectation> to increase.
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