Let $R > 0 $, $z \in \mathbb{C}, \ f : D(z,R) \rightarrow \mathbb{C} $.
$Re(f) \ $ and $Im(f) \ $ are $C^{1} $ on $D(z,R) \ $. Then f is complex differentiable in $z$ if and only if
$$ \lim_{r \rightarrow 0^{+}} \frac{1}{\pi r^{2}}\int_{∂D(z,R)}f(\eta)d \eta = 0$$
How to solve it ?
First of all, "holomorphic in [at] $z$" doesn't really make sense. Holomorphy is a condition for functions on open sets. I assume you mean "complex differentiable at $z$". (Also, you're using $z$ to mean two different things and mixing $r$ and $R$.)
First, assume that $f'(z)$ exists. Then $f(w) = f(z) + (z-w)f'(z) + o(|z-w|)$. Hence $$ \int_{|w-z|=r} f(w)\,dw = \int_{|w-z|=r} (f(z) + (z-w)f'(z) + o(r))\,dw = o(r^2) $$ since the first two terms vanish and the third term picks up an extra $r$ from the length of the curve. Thus the requested limit is $0$ as $r \to 0$.
Conversely, split $f$ in its real and imaginary parts, $f = u+iv$ and put $w=x+iy$. By Green's formula: \begin{align} \int_{|w-z|=r} f(w)\,dw &= \int_{|w-z|=r} (u+iv)(dx+i\,dy) \\ &= \int_{|w-z|=r} u\,dx -v\,dy + i\int_{|w-z|=r} v\,dx + u\,dy \\ &= \iint_{|w-r|\le r} (-v'_x-u'_y)\,dx\,dy + i \iint_{|w-r|\le r} (u'_x-v'_y)\,dx\,dy. \end{align} Assume that $-v'_x(z)-u'_y(z) = A \neq 0$. Then (by continuity of the partials), if $r$ is small, the integral $$\iint_{|w-r|\le r} (-v'_x-u'_y)\,dx\,dy$$ is very close to $Ar^2$, which means that the requested limit can't exist (fill in the details!). Hence $v'_x(z) = -u'_y(z)$ and a similar argument shows that $u'_x(z)=v'_y(z)$. This means that Cauchy-Riemann's equations are satsified at $z$ and since $f\in C^1$ this is enough to guarantee that $f$ is complex differentiable at $z$.