= Solution
Start the <planar Brownian motion> $X_t+iY_t$ at $-x$ and stop at $\tau=\inf\{t:X_t=0\}$. The complex exponential is holomorphic with nonzero derivative. By <conformal invariance of planar Brownian motion>, the process $e^{X_t+iY_t}$, run on its <conformal Brownian clock>
$$
A_t=\int_0^{t\wedge\tau}e^{2X_s}\,ds,
$$
is planar <Brownian motion> started at $\varepsilon=e^{-x}$, until its first exit from the unit disk. The clock is strictly increasing before $\tau$, and $A_\tau\leq\tau<\infty$ almost surely. It is legitimate to use the exponential although it is not globally injective: the Brownian coordinate brackets depend locally on the nonzero derivative, and the clock construction gives the stopped planar Brownian law. Before $\tau$, $|e^{X_t+iY_t}|<1$; at $\tau$ it equals one. Hence the disk exit time in the new clock is exactly $A_\tau$.
The continuous lifted argument of this path, starting from angle zero, is $Y_t$. Its total argument at exit is therefore $Y_\tau$, which has distribution $xC$ by part (b). If the winding count records only completed integer turns, it differs from $Y_\tau/(2\pi)$ by a remainder $R_\varepsilon$ with $|R_\varepsilon|\leq1$. If fractional turns are retained, that remainder is zero. In either convention, since $\log\varepsilon=-x$,
$$
\frac{\theta_\varepsilon}{\log\varepsilon}
=-\frac{Y_\tau}{2\pi x}-\frac{R_\varepsilon}{x}.
$$
The first term has distribution $-C/(2\pi)$ for every $x$, and the second tends to zero deterministically in absolute value as $x\to\infty$. Symmetry of the <Cauchy distribution> removes the minus sign, so <Slutsky theorem> gives the <small-radius Brownian winding law>:
$$
\boxed{\frac{\theta_\varepsilon}{\log\varepsilon}\ \xrightarrow[\varepsilon\downarrow0]{d}\ \frac{C}{2\pi}.}
$$
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