= Solution
Suppose $u_1,u_2$ are solutions and put $w=u_1-u_2$. Then $\Delta w=0$ and $w$ satisfies the homogeneous <Robin boundary condition>
$$
(1-\lambda)w+\lambda\frac{\partial w}{\partial n}=0.
$$
For $0<\lambda<1$, <Green's first identity> gives
$$
\int_V|\nabla w|^2\,dV
=\int_Sw\frac{\partial w}{\partial n}\,dS
=-\frac{1-\lambda}{\lambda}\int_Sw^2\,dS\leq0.
$$
The left side is nonnegative, so both sides vanish and $w=0$. For $\lambda=0$, the <Dirichlet boundary condition> $w=0$ on $S$ and the same identity again imply $w=0$. The solution is therefore unique for every $0\leq\lambda<1$.
For a regular spherically symmetric solution, the <Laplacian> equation is
$$
\frac1{r^2}\frac d{dr}\left(r^2u'\right)=6.
$$
Regularity at the origin gives $u'=2r$, hence $u=r^2+C$. Applying the boundary condition at $r=b$ yields
$$
\boxed{u(r)=r^2-b^2-\frac{2\lambda b}{1-\lambda}},
\qquad 0\leq\lambda<1.
$$
At $\lambda=1$ the <Neumann boundary condition> would require $u'(b)=2b=0$, which is impossible for $b>0$. Equivalently, its compatibility condition would require $\int_V6\,dV=0$.
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