One could work here with the PDF $$f_{\bf Z}(x,y)={1\over\pi}\qquad\bigl(0\leq\sqrt{x^2+y^2}\leq a^2\bigr)\ ,$$ whereby the random variable ${\bf Z}=(X,Y)$ denotes the exact spot where the stone falls. This then would lead to the expected value $$E(R)={1\over\pi}\int_{B(a,{\bf 0})}\sqrt{x^2+y^2}\>{\rm d}(x,y)$$ of the distance $R:=|{\bf Z}|$. The integral $(1)$ has to be computed via a transformation to polar coordinates. The following setup circumvents this and brings only the variable $R$ into the game.
The distance $R$ of the stone from the center is a random variable supported on the interval $[0,a]$, where $a$ denotes the radius of the pond. For a given $r\in[0,a]$ the cumulated probability that $R\leq r$ amounts to $$F(r)={\pi r^2\over \pi a^2}\ .$$ Therefore the probability density of the random variable $R$ is given by $$f_R(r)=F'(r)={2r\over a^2}\ ,$$ and the expected distance computes to $$E(R)=\int_0^a r\>f_R(r)\>dr={1\over a^2}\int_0^a 2r^2\>dr={2\over3}a\ .\tag{1}$$