The heuristics behind that answer is the following. Consider the probability $P(r)$ that this distance is at least $r$. If equals $\int_Q (1-U(x,r))^N\\,dx$ where $U(x,r)=\int_{B(x,r)}f(x)\\,dx$. Now, $U(x,r)\approx \pi f(x)r^2$, and $(1-U)^N\approx e^{-NU}$, so the expectation in question is
$$
\int_Q \int_0^\infty (1-U(x,r))^N\\,dr\\,dx\approx\int_Q \int_0^\infty e^{-\pi Nf(x)r^2}\\,dr\\,dx=
\frac 1{2\sqrt N} \int_Q f(x)^{-1/2}\\,dx
$$
To make this all a rigorous large $N$ asymptotics, we need some conditions on $f$. Of course, if $f^{-1/2}$ is not integrable, the result is meaningless. However, if $f^{-1/2}\in L^1$, then $f^{-1/4}\in L^2$, so $g=M(f^{-1/4})\in L^2$ ($M$ stands for the Hardy-Littlewood maximal function) whence, by Holder, $U(x,r)\ge \pi g(x)^{-4} r^2$ and, finally $(1-U(x,r))\le e^{-\pi g(x)^{-4} r^2}$ giving us the integrable majorant. I'm cheating a bit because we should be a little bit more careful with $x$ near the boundary of the square and with large $r$, of course, but it all works out fine and I leave those details to you to figure out :).