= Solution
Set $F(w)=(\phi(w)-\phi(0))/w$ for $w\ne0$. The singularity is removable, with $F(0)=\phi'(0)$. Injectivity of the <conformal map> makes $F(w)\ne0$ away from zero, and conformality makes $F(0)\ne0$ as well. Since the disc is simply connected, $F$ has a <holomorphic logarithm>. Therefore $\psi=\log|F|$ is a <harmonic function>, including at zero, and $\psi(0)=\log|\phi'(0)|$.
The <mean value property for harmonic functions> gives, for every $0<r<1$,
$$
\log|\phi'(0)|=\frac1{2\pi}\int_0^{2\pi}\log|\phi(re^{i\theta})-\phi(0)|\,d\theta-\log r.
$$
For general $D$, boundary values mean radial limits, rather than a continuous extension of $\phi$ to every point of the circle. The <boundary logarithmic mean of a univalent function> justifies taking $r\uparrow1$: the <Koebe distortion theorem> bounds $|F|$ below by $|\phi'(0)|/4$, while the standard integral-mean bound for a <univalent function>, $\sup_{r<1}\int|\phi(re^{i\theta})|^p d\theta<\infty$ for $0<p<1/2$, gives <uniform integrability> of its positive logarithm. Radial limits exist almost everywhere, and passage to the integral follows. Thus, writing $dw=d\theta$ for arc length on the unit circle,
$$
\boxed{\log|\phi'(0)|=\frac1{2\pi}\int_{\partial\mathbb D}\log|\phi(w)-\phi(0)|\,dw.}
$$
The expression on the left is the logarithm of the <conformal radius> of $D$ at $z=\phi(0)$.
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