= Solution
An <isotropic tensor> has unchanged components under every rotation: for rank four, $A_{ijkl}=R_{ia}R_{jb}R_{kc}R_{ld}A_{abcd}$. Orthogonality gives $R_{ia}R_{ja}=\delta_{ij}$, so each product of paired <Kronecker deltas> is invariant. Every linear combination of $\delta_{ij}\delta_{kl}$, $\delta_{ik}\delta_{jl}$ and $\delta_{il}\delta_{jk}$ is therefore isotropic, for arbitrary coefficients.
For the integral, differentiation away from the origin gives
$$
\partial_k\partial_l\frac1r=\frac{3x_kx_l}{r^5}-\frac{\delta_{kl}}{r^3}.
$$
After multiplication by $x_ix_j$, this behaves as $O(r^{-1})$ and is locally integrable in three dimensions. Interpret the integral as the limit with a small central ball removed. A rotation preserves the integration ball and the radial kernel, so the resulting $B$ is an <isotropic tensor integral>. The symmetries $i\leftrightarrow j$ and $k\leftrightarrow l$ force the two cross-pair coefficients equal:
$$
B_{ijkl}=A\delta_{ij}\delta_{kl}+B(\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk}).
$$
Taking the trace in $k,l$ uses $\nabla^2(r^{-1})=0$ off the origin and gives $3A+2B=0$. For the other contraction, the displayed Hessian gives $\sum_{ij}x_ix_j\partial_i\partial_j(r^{-1})=2/r$. Thus
$$
3A+12B=\int_{r<a}\frac2r\,dV=8\pi\int_0^a r\,dr=4\pi a^2.
$$
Solving gives $B=2\pi a^2/5$ and $A=-4\pi a^2/15$. Therefore \b[the requested tensor is]
$$
\boxed{B_{ijkl}=\frac{2\pi a^2}{15}\left(-2\delta_{ij}\delta_{kl}+3\delta_{ik}\delta_{jl}+3\delta_{il}\delta_{jk}\right).}
$$
This is the <quadratically weighted inverse-radius Hessian integral>. The origin creates no extra contribution to this weighted integral; even the distributional point term in the unweighted Hessian is annihilated by $x_ix_j$.
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