I know how to prove this in general (I have a proof for Poisson formula I did), however, I have not been able to prove this way. Please help.
Assume the theorem for $R=1$ deduce statement for any $R>0$. That is,
Suppose that $u(z)$ is harmonic on $B(0,R)$ and continuous for $\overline{B(0,R)}$. Then $$u(a) = \frac {1}{2 \pi} \int_{|z| = R} \frac {R^2 - |a|^2}{|z-a|^2}u(z)d\theta.$$
How do I get that from the assumption?
Suppose that $u(z)$ is harmonic on $B(0,R)$ and continuous for $\overline{B(0,R)}$. Then $$v :\overline{ B(0,1)} \to \mathbb R, \ \text{where }v(x) = u(Rx)$$ is harmonic in $B(0,1)$ and continuous in $\overline{ B(0,1)}$. Then for all $b\in B(0,1)$,
$$v(b) = \frac {1}{2 \pi} \int_{|w| = 1} \frac {1- |b|^2}{|w-b|^2}v(w)d\theta.$$
which is $$u(Rb) = \frac {1}{2 \pi} \int_{|w| = 1} \frac {1- |b|^2}{|w-b|^2}u(Rw)d\theta.$$ Writing $a = Rb$ and $z = Rw$, then $$\begin{split} u(a) &= \frac {1}{2 \pi} \int_{|z| = R} \frac {1- |a/R|^2}{|z/R-a/R|^2}u(z)d\theta \\ &=\frac {1}{2 \pi} \int_{|z| = R} \frac {R^2- |a|^2}{|z-a|^2}u(z)d\theta . \end{split}$$