1. Given potential inside the spherical ball

   $V(r)=2ar^2+b$

   where $a,b$ are constants.

2. Use Poisson's equation

   For electrostatics in a medium of permittivity $\varepsilon$,

   $\nabla^2 V=-\frac{\rho}{\varepsilon}$

   Since the potential depends only on $r$, in spherical coordinates:

   $\nabla^2 V=\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{dV}{dr}\right)$

3. Differentiate the potential

   $\frac{dV}{dr}=\frac{d}{dr}(2ar^2+b)=4ar$

   Then,

   $r^2\frac{dV}{dr}=r^2(4ar)=4ar^3$

   Differentiate again:

   $\frac{d}{dr}(4ar^3)=12ar^2$

   Therefore,

   $\nabla^2 V=\frac{1}{r^2}(12ar^2)=12a$

4. Apply Poisson's equation

   $12a=-\frac{\rho}{\varepsilon}$

   so

   $\rho=-12a\varepsilon$

5. Compare with the given form

   The charge density is given as

   $\rho=-\lambda a\varepsilon$

   Comparing,

   $-\lambda a\varepsilon=-12a\varepsilon$

   hence,

   $\lambda=12$

6. Final answer

   $\boxed{12}$

The derived answer matches the stored correct answer.