= Solution
By <conformal invariance of extremal length>, work first on $1<|w|<R$. For a <conformal metric> $\rho$, let $L_\rho$ be the infimum of lengths of curves joining the two circles, and $A_\rho$ its area. Each radial segment is admissible, so the <Cauchy-Schwarz inequality> gives, for every angle,
$$
L_\rho^2\leq\left(\int_1^R\rho(se^{i\theta})ds\right)^2\leq\log R\int_1^R\rho(se^{i\theta})^2s\,ds.
$$
Integrating in angle gives $2\pi L_\rho^2\leq A_\rho\log R$. Conversely, $\rho(w)=1/|w|$ has shortest joining length $\log R$ and area $2\pi\log R$, attaining the bound. Thus the <conformal modulus of an annulus> gives
$$
\boxed{d_\Omega(E,F)=\frac{\log R}{2\pi},\qquad\rho_\Omega(z)=c\left|\frac{\psi'(z)}{\psi(z)}\right|,\quad c>0}.
$$
The second formula is the pullback of $c/|w|$ by $\psi$. Pullback preserves curve lengths and metric area, so every displayed positive multiple is an extremal metric in $\Omega$.
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