The minimum work equals a change in the energy of electric field. By comparing the electric field of the initial and final configurations, one can conclude that the energy increment is equal to $W_{2}-W_{1}$, where $W_{1}$ is the energy of the field in the dielectric sphere with inner and outer radii $R_{2}$ and $R_{3}$ (the field is due to the charge $Q$ located at the center) and $W_{2}$ is the field energy in the «empty» space between the spheres with radii $R_{1}$ and $R_{3}$ (the field is due to the charge $Q$ located at the common centre of the spheres). It is convenient to find $W_{1}$ and $W_{2}$ as the energies of the corresponding spherical capacitors $C_{1}$ and $C_{2}$ with the charges equal to $Q$ on their plates.
Let us determine $C_{2}$ and $W_{2}$. A voltage across a spherical capacitor with plate radii $R_{1}$ and $R_{3}$ is
\[
U=\left(k \frac{Q}{R_{1}}-k \frac{Q}{R_{3}}\right)-0=k Q \frac{R_{3}-R_{1}}{R_{1} R_{3}}, \quad \text { where } \quad k=\frac{1}{4 \pi \varepsilon_{0}}
\] The capacitance is
\[
C_{2}=\frac{Q}{U}=\frac{R_{1} R_{3}}{k\left(R_{3}-R_{1}\right)}
\] and the energy is
\[
W_{2}=\frac{Q^{2}}{2 C_{2}}=\frac{k Q^{2}\left(R_{3}-R_{1}\right)}{2 R_{1} R_{3}}
\] Similarly,
\[
C_{1}=\frac{\varepsilon R_{2} R_{3}}{k\left(R_{3}-R_{2}\right)}, \quad W_{1}=\frac{k Q^{2}\left(R_{3}-R_{2}\right)}{2 \varepsilon R_{2} R_{3}}
\] The required work is
\[
A=W_{2}-W_{1}=\frac{k Q^{2}}{2 R_{3}}\left(\frac{R_{3}-R_{1}}{R_{1}}-\frac{R_{3}-R_{2}}{\varepsilon R_{2}}\right)
\]