This question is very similar to a previous one I asked, but with a substantial difference.
Let $r=\sqrt{x^2+y^2+z^2}$. Consider the function $f:\mathbb{R^3}\rightarrow\mathbb{R} $ such that $$f(x,y,z)=\begin{cases} r & \text{if $r\leq 1$} \\ 0 & \text{if $r>1$} \end{cases} $$
Is there a function $\phi$ such that $$\Delta\phi=f$$ where $\Delta\phi=\frac{\partial ^2\phi}{\partial x^2}+\frac{\partial ^2\phi}{\partial y^2}+\frac{\partial ^2\phi}{\partial z^2}$?
If so, is the solution unique (up to a constant)? And how do I find it?
As mentioned in the comments, there are solutions of $\Delta g=0$, and these can be added to any solution $\phi$ of yours to come up with another solution of $\Delta \phi=f$.
You can expect to find a radial solution $F(r)$ because the right side of your equation is radial, and because the Laplacian has a nice form for radial functions. Assuming you're working in $\mathbb{R}^3$, $$ \Delta F(r) = F''(r)+\frac{2}{r}F'(r). $$ You have a function $H(r)$ that is $r$ for $0 \le r \le 1$ and is $0$ otherwise, and you want to solve $\Delta F(r)=H(r)$. Your function $H$ is discontinuous, but that doesn't matter much. $$ r^2F''(r)+2rF'(r) = r^2H(r) \\ (r^2F')'=r^2H(r) \\ r^2F'(r)= \int_{0}^{r}s^2H(s)ds \\ F(r) = \int_{0}^{r}\frac{1}{t^2}\int_{0}^{t}s^2H(s)dsdt. $$ For $r \le 1$, the solution is $$ F(r) = \int_{0}^{r}\frac{1}{t^2}\int_{0}^{t}s^3dsdt= \frac{r^3}{12} $$ For $r \ge 1$, the solution is $$ F(r) = \frac{1}{12}+\int_{1}^{r}\frac{1}{t^2}\frac{1}{4}dt=\frac{1}{12}-\left.\frac{1}{4t}\right|_{t=1}^{r} = \frac{1}{12}-\frac{1}{4}+\frac{1}{4r} $$