Time for a particle undergoing brownian motion to reach a point in a volume

First remark, the average hitting time is finite because the volume is finite. None of what I write would make sense in an infinite system.

Let us consider that the target is a ball of radius $a$ at the center of the sphere and let us call $T(\vec r)$ the average hitting time for a Brownian particle starting at position $\vec r$ from the origin. $T(\vec r)$ depends only on $r=\|\vec r\|$. Consider now a Brownian motion $\vec B$ during a short time $\mathrm dt$ and compute the variation of the hitting time $$\mathrm dt=T(\vec r)-T(\vec r+\vec B_{\mathrm dt}).$$ Use the Taylor expansion in spherical coordinates of the right-hand side $$\mathrm dt=-\vec\nabla T(\vec r)\cdot\vec B_{\mathrm dt}-\frac12\vec B_{\mathrm dt}\cdot H_T(\vec r)\cdot \vec B_{\mathrm dt}$$ where $H_T$ is the Hessian matrix of $T$ which contains only one non-zero element equal to $\frac1{r^2}\partial_r(r^2\partial_r T(r))$ on the diagonal. Taking the average over all realisations of the Brownian motion, one gets $$\mathrm dt=-\frac12\frac1{r^2}\partial_r\left(r^2\partial_r T(r)\right) \; 2D\mathrm dt,$$ or in a simpler form, with $\Delta_r$ denoting the spherical Laplacian, $$D\Delta_r T(r)=-1.\tag{1}$$ This form is quite general for the average hitting time. It is actually the Fokker-Planck equation in which we have replaced the time derivative by $-1$.

Let us solve (1). We have $$T(r)=-\frac{r^2}{6D}+\frac{A}r+B.$$ $A$ and $B$ are constants. We must have $T(a)=0$. The second condition depends on the boundary at $r=R$. Let us consider that the boudary is reflecting, so the particles bounce on the sphere and continue to diffuse inside. This is a von Neumann boundary condition, that translates mathematically into $\left.\partial_rT(r)\right\rvert_{r=R}=0$. This defines $A=-R^3/3D$. With $T(a)=0$ we find $B=a^2/6D+R^3/3Da$. Finally $$T(r)=\frac{a^2-r^2}{6D}+\frac{R^3}{3D}\left(\frac1a-\frac1r\right).$$

If the starting point is uniformally distributed inside the sphere of radius $R$, the average is $$ \int_a^R\frac{3r^2}{R^3-a^3}T(r)\mathrm dr=\frac{(R-a)^2}{15Da}\frac{5R^3+6 R^2a + 3 Ra^2 + a^3}{R^2+Ra+a^2}.$$

Therefore, for $a\ll R$ we get $$\left\langle T\right\rangle\approx\frac{R^3}{3Da}.$$ The average hitting time scales like $R^3$, so is actually proportional to the volume and inversely proportional to the radius of the target.