= Solution
On an outward-oriented sphere of radius $R>0$, the <electric field> is normal with constant magnitude $A/R^2$. Thus its <flux> is
$$
\boxed{\iint_{S_R}E\cdot d\boldsymbol S=\frac A{R^2}(4\pi R^2)=4\pi A=\frac Q{\epsilon_0}.}
$$
It is independent of $R$. There is no contradiction with <8/b/i/solution>: the classical <divergence theorem> cannot be applied to a full ball while pretending that $E$ is smooth at its center. On a punctured ball the outer flux and the inward-normal inner flux cancel, in agreement with zero ordinary <divergence> there.
The three-dimensional <Dirac delta distribution> $\delta^{(3)}(r)$ is supported at the origin and satisfies the sampling identity
$$
\int_{\mathbb R^3}\delta^{(3)}(r)\varphi(r)\,dV=\varphi(0)
$$
for smooth compactly supported test functions $\varphi$. It is not an ordinary function with a finite value at zero; its total integral is one, and translating it moves the point at which the test function is sampled.
The globally valid source equation is
$$
\boxed{\nabla\cdot E=4\pi A\delta^{(3)}(r)=\frac Q{\epsilon_0}\delta^{(3)}(r).}
$$
For a direct weak verification, excise a ball of radius $\varepsilon$ and integrate $-E\cdot\nabla\varphi$ outside it. Since the ordinary <divergence> vanishes there, the result is $A\int_{S^2}\varphi(\varepsilon\hat r)\,d\Omega\to4\pi A\varphi(0)$. This is precisely the displayed <distributional divergence> and restores the <divergence theorem> with the <point charge> included.
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