Laplacian defined as an integration over a 3D ball. In the book "Theory of unitary symmetry" by Rumer and Fet (see a piece of text at the bottom of this post) there is a proof which uses the following result (below I will translate as close as possible to the original, equation enumeration is changed, some irrelevant to this post pieces are skipped (replaced by ellipsis):

... In order to prove the first of these relations let us introduce the following operator
$$\Delta_{\varepsilon}\psi = \frac{3}{4\pi\varepsilon^2}\iiint\limits_{K_{x,\varepsilon}}\psi\mathrm{d}v - \psi(x),\label{eq:1a}\tag{1a}$$ where
$K_{x, \varepsilon}$ - is a ball with radius $\varepsilon$ and
centered at the point $x$, and $\mathrm{d}v$ - volume element.
...
The last step is to notice that $$ \Delta \psi(x) = \lim_{\varepsilon\to 0} \frac{10}{\varepsilon^3} \Delta_{\varepsilon}\psi.\label{eq:2}\tag{2} $$

Then there is a hint how to prove this by expanding $\psi$ in Taylor series and then integrate. But problem is I couldn't get the result. More over, I think that formula is incorrect because the class of the first term in \ref{eq:1a} is $O(\varepsilon)$ while the second term is in $O(1)$. Am I wrong? And how to prove  \ref{eq:2} if not.
My try
Rewriting eq.\ref{eq:1a} with a correction:
$$\Delta_{\varepsilon}\psi = \frac{3}{4\pi\varepsilon^2}\iiint\limits_{K_{x, \varepsilon}}(\psi(x') - \psi(x))\mathrm{d}v',\label{eq:1b}\tag{1b}
$$
I have proved that eq.\ref{eq:2} holds for it. But eqs.\ref{eq:1a}-\ref{eq:1b} are not equivalent.
Original text

 A: You are right the equations should be, where I've chosen the ball to be centered at the origin
$$\Delta_\epsilon\psi=\frac{3}{4\pi \epsilon^3}\int \psi(x) \,dv-\psi(0),\quad \Delta\psi=\lim_{\epsilon\rightarrow 0}\frac{10}{\epsilon^2}\Delta_\epsilon\psi$$
To prove it we start as you did,
$$\Delta_\epsilon\psi=\frac{3}{4\pi \epsilon^3}\int \psi(x) -\psi(0)\,dv$$
Then we Taylor expand around $0$ to 2nd order, which gives terms proportional to $x$, $xy$ and $x^2$, but due to symmetry the $x$ and $xy$ kind of terms integrate to zero over the ball thus we have,
$$\Delta_\epsilon\psi=\frac{3}{4\pi \epsilon^3}\left[\frac{1}{2}\frac{\partial^2\psi}{\partial x^2}\int x^2\,d v+\frac{1}{2}\frac{\partial^2\psi}{\partial y^2}\int y^2\,d v+\frac{1}{2}\frac{\partial^2\psi}{\partial z^2}\int z^2\,d v\right]+O(\epsilon^3)$$
where all derivatives are evaluated at the origin.The integrals all give the same value
$$\int x^2\,dv=\frac{1}{3}\int x^2+y^2+z^2\,dv=\frac{1}{3}4\pi\int_o^\epsilon r^4\,dr=\frac{4\pi\epsilon^5}{15}$$
If we insert this
$$\Delta_\epsilon\psi=\frac{3}{4\pi \epsilon^3}\frac{1}{2}\frac{4\pi\epsilon^5}{15}\left[\frac{\partial^2\psi}{\partial x^2}+\frac{\partial^2\psi}{\partial y^2}+\frac{\partial^2\psi}{\partial z^2}\right]+O(\epsilon^3)=\frac{\epsilon^2}{10}\Delta\psi+O(\epsilon^3)$$
Finally we take the limit
$$\lim_{\epsilon\rightarrow 0}\frac{10}{\epsilon^2}\Delta_\epsilon\psi=\lim_{\epsilon\rightarrow 0}\left[\Delta\psi+O(\epsilon)\right]=\Delta\psi$$
