= Solution
For $0<|z|<R$, set $w=R^2/\bar z$, which lies outside the <closed disk>. On $|\zeta|=R$, direct algebra gives the <Poisson kernel> identity
$$
\operatorname{Re}\frac{\zeta+z}{\zeta-z}=\frac{\zeta}{\zeta-z}-\frac{\zeta}{\zeta-w}.
$$
Using $d\theta=d\zeta/(i\zeta)$, the proposed <Poisson formula for a holomorphic function> becomes
$$
\frac1{2\pi i}\oint_{|\zeta|=R}f(\zeta)\left[\frac1{\zeta-z}-\frac1{\zeta-w}\right]d\zeta=f(z).
$$
The first term is $f(z)$ by the <Cauchy integral formula>; the second vanishes by the <Cauchy integral theorem>, since its pole lies outside the disk. At $z=0$ the kernel is one, and the same identity is the <mean value property for holomorphic functions>. Therefore
$$
\boxed{f(z)=\frac1{2\pi}\int_0^{2\pi}f(Re^{i\theta})\operatorname{Re}\frac{Re^{i\theta}+z}{Re^{i\theta}-z}\,d\theta\qquad(|z|<R)}.
$$
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