Let $\,r\,$ be the radius of a spherical surface with uniform surface charge density $\,\sigma\boldsymbol >0\,$ so with charge $\,q\boldsymbol=4\,\pi\,r^2\sigma\boldsymbol >0$. Then the field inside this spherical surface is zero while the field outside is the same as if the charge $\,q\boldsymbol >0\,$ is concentrated on the center of the sphere so directed outwards.
For the case of a point in the interior of a solid sphere (ball) with uniform volume charge density $\,\rho\boldsymbol >0\,$ think that the field due to the sum (integral) of the spherical shells beyond the reference point have zero contribution while the field due to the sum (integral) of the spherical shells below the reference point have a contribution directed outwards as if the charge of the solid sphere below the reference point is concentrated on the center. Of cource if $\,\rho\boldsymbol <0\,$ then the field is directed inwards.
By an other argument : if a closed smooth surface like a sphere of radius $\,R\,$ is oriented with its normal $\,\mathbf n\,$ directed outwards then by Gauss Law for the electric flux $\,\Phi\boldsymbol=Q/\epsilon_0\,$ where $\,Q\,$ the enclosed electric charge. But $\,\Phi\boldsymbol=4\,\pi\,R^2 E\boldsymbol=Q/\epsilon_0\,$ where $\,E\,$ the algebraic projection of $\,\mathbf E\,$ on $\,\mathbf n\,$ at every point on the surface $\,E\boldsymbol=\mathbf E\boldsymbol\cdot\mathbf n$. So $\,E\,$
must have the sign of $\,Q$.