= Solution
Put $\mu=\mu^0$, $\lambda=\kappa^0-2\mu/3$, and $L=\lambda+2\mu=\kappa^0+4\mu/3$. Assume positive <bulk modulus> and <shear modulus>. For a unit vector $\xi$, direct contraction of the isotropic <elastic stiffness tensor> gives the <acoustic tensor> and its inverse:
$$
K(\xi)=\mu I+(L-\mu)\xi\otimes\xi,
\qquad K^{-1}(\xi)=\frac1\mu(I-\xi\otimes\xi)+\frac1L\xi\otimes\xi.
$$
These are its transverse and longitudinal <orthogonal projections>. Symmetrizing the definition of the <directional elastic strain Green operator> consequently gives
$$
\widetilde\Gamma_{ijkl}(\xi)
=\frac1{4\mu}\bigl(\delta_{jk}\xi_i\xi_l+\delta_{ik}\xi_j\xi_l+\delta_{jl}\xi_i\xi_k+\delta_{il}\xi_j\xi_k\bigr)
+\left(\frac1L-\frac1\mu\right)\xi_i\xi_j\xi_k\xi_l.
$$
Substitution into the spherical representation is already the isotropic specialization. We can also evaluate its angular integrals as <distributions>. For $r=|x|>0$, direct integration with the polar axis parallel to $x$ gives
$$
\int_{S^2}\delta(\xi\cdot x)\,dS=\frac{2\pi}{r},
\qquad\int_{S^2}|\xi\cdot x|\,dS=2\pi r.
$$
The first identity holds as a locally integrable <distribution>. Since the second derivative of $|s|$ is $2\delta(s)$, taking two derivatives in the first identity and four in the second gives
$$
\int_{S^2}\xi_i\xi_j\delta''(\xi\cdot x)\,dS=2\pi\partial_i\partial_j(r^{-1}),
\qquad
\int_{S^2}\xi_i\xi_j\xi_k\xi_l\delta''(\xi\cdot x)\,dS=\pi\partial_i\partial_j\partial_k\partial_l r.
$$
Thus an explicit complete <isotropic elastic strain Green kernel> is
$$
\begin{aligned}
\Gamma_{ijkl}(x)
={}&-\frac1{16\pi\mu}\bigl(\delta_{jk}\partial_i\partial_l+\delta_{ik}\partial_j\partial_l+\delta_{jl}\partial_i\partial_k+\delta_{il}\partial_j\partial_k\bigr)r^{-1}\\
&-\frac1{8\pi}\left(\frac1L-\frac1\mu\right)\partial_i\partial_j\partial_k\partial_l r.
\end{aligned}
$$
All derivatives here are <distributional derivatives>, which retain the contribution at the origin.
For the requested contraction, $\widetilde\Gamma_{ijkk}=\xi_i\xi_j/L$, or equivalently trace the complete kernel and use $\Delta r=2/r$. Therefore
$$
\boxed{\Gamma_{ijkk}(x)=-\frac1{4\pi L}\partial_i\partial_j\frac1{|x|}.}
$$
More explicitly, its <distribution> form is
$$
\Gamma_{ijkk}
=\frac{\delta_{ij}}{3L}\delta(x)
+\frac1{4\pi L}\operatorname{pv}\frac{\delta_{ij}|x|^2-3x_i x_j}{|x|^5},
$$
where the <Cauchy principal value> uses spherical exclusions about the origin. The coefficient of the <Dirac delta> can be checked by integrating by parts over a punctured ball: $\partial_i\partial_j(r^{-1})$ has contact term $-4\pi\delta_{ij}\delta/3$. Taking the trace, the principal-value part has zero trace and $\Delta(r^{-1})=-4\pi\delta$, so
$$
\boxed{\Gamma_{iikk}(x)=\frac1L\delta(x).}
$$
Keeping only the ordinary kernel away from the origin would incorrectly give zero for this double trace.
Now extend the <elastic polarization> by zero outside $V$. Since the <shear modulus> is unchanged, the stiffness contrast acts only on the spherical <strain>:
$$
\tau_{ij}(x)=(C-C^0)_{ijkl}\varepsilon_{kl}(x)
=\bigl(\kappa(x)-\kappa^0\bigr)\varepsilon_{kk}(x)\delta_{ij}.
$$
Hence $\tau=(\kappa-\kappa^0)\operatorname{tr}\varepsilon$ is its scalar coefficient. Relative to the same external loading, the <elastic strain Green operator> gives $\varepsilon=\varepsilon^0-\Gamma*(\tau I)$. Taking the trace and using the contact identity reduces this nonlocal equation to
$$
\varepsilon_{kk}(x)=\varepsilon^0_{kk}(x)-\frac{\kappa(x)-\kappa^0}{L}\varepsilon_{kk}(x).
$$
Solving the scalar equation gives the <constant-shear bulk-modulus inclusion> formula
$$
\boxed{\varepsilon_{kk}(x)=\frac{3\kappa^0+4\mu^0}{3\kappa(x)+4\mu^0}\varepsilon^0_{kk}(x).}
$$
Using the single contraction instead of the double trace gives
$$
\varepsilon_{ij}(x)=\varepsilon^0_{ij}(x)+\frac1{4\pi L}\partial_i\partial_j
\int_V\frac{(\kappa(x')-\kappa^0)\varepsilon_{kk}(x')}{|x-x'|}\,dx'.
$$
Finally eliminate the actual trace using the scalar formula:
$$
\boxed{\varepsilon_{ij}(x)=\varepsilon^0_{ij}(x)+\frac1{4\pi}\partial_i\partial_j
\int_V\frac{3(\kappa(x')-\kappa^0)}{3\kappa(x')+4\mu^0}
\frac{\varepsilon^0_{kk}(x')}{|x-x'|}\,dx'.}
$$
The reference trace $\varepsilon^0_{kk}$ in this last formula is essential and is what the original PDF prints; the converted TeX loses its superscript. The second derivatives are interpreted distributionally, or with their principal-value and contact terms explicitly included. With ordinary regularity they give the corresponding pointwise formula at regular interior points.
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