= Solution
For a spherically symmetric <function> in three dimensions, the <radial Laplacian> is $\nabla^2u=r^{-2}(r^2u')'$. On $0<r\leq a$, integrate $(r^2u')'=r^2$ twice to obtain
$$
u'=\frac r3+\frac{A}{r^2},\qquad u(r)=\frac{r^2}{6}-\frac Ar+B.
$$
The region includes the origin. A classical solution must be regular there, so $A=0$: the $1/r$ term is singular and is not a solution of the stated full-ball problem. Hence the complete family on the ball is
$$
\boxed{u(r)=\frac{r^2}{6}+B.}
$$
The boundary cases below use this regular family. Allowing punctured-ball solutions would change the problem and the compatibility calculation.
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