= Solution
The <concentric normalization of an eccentric annulus> here uses
$$
T(z)=\frac{z-1/2}{1-z/2}.
$$
This <Möbius transformation> preserves the unit circle and sends the inner circle's real endpoints $0,4/5$ to $-1/2,1/2$. Its image circle is symmetric about the real axis and hence is $|w|=1/2$. Equivalently, expanding $|T(z)|^2=1/4$ gives $|z|^2=(4/5)\operatorname{Re}z$, precisely the inner-circle equation. Thus $T$ sends the domain to $1/2<|w|<1$. <Conformal invariance of harmonic measure> and <3/a/i> give
$$
\boxed{\omega_\Omega(z,\mathbb T)=\frac{\log(2|T(z)|)}{\log2}=1+\frac{\log\left|\frac{z-1/2}{1-z/2}\right|}{\log2}}.
$$
The zero of $T$ lies in the removed inner disk, so this logarithm is harmonic throughout $\Omega$.
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