= Solution
In electrostatics, <Gauss's law> is $\oint_{\partial\Omega}\mathbf E\cdot\mathbf n\,dS=Q_{\rm enclosed}/\epsilon_0$, or locally $\nabla\cdot\mathbf E=\rho_e/\epsilon_0$, where $\rho_e$ is charge density and $\epsilon_0$ is <vacuum permittivity>. <Spherical symmetry> makes the <electric field> radial, and Gauss's law gives
$$
\boxed{\mathbf E(r)=\begin{cases}0,&r<a,\\ \dfrac{Q}{4\pi\epsilon_0r^2}\mathbf e_r,&a<r<b,\\0,&r>b.\end{cases}}
$$
Choose the additive constant of the <electrostatic potential> by $\phi(\infty)=0$. Using $\mathbf E=-\nabla\phi$ and continuity of potential across each charged shell gives
$$
\boxed{\phi(r)=\begin{cases}\dfrac{Q}{4\pi\epsilon_0}(1/a-1/b),&r<a,\\ \dfrac{Q}{4\pi\epsilon_0}(1/r-1/b),&a<r<b,\\0,&r>b.\end{cases}}
$$
The field jumps at the shell surfaces, so the displayed formulas are for the requested open regions. The potential difference of the <spherical capacitor> is $\Delta\phi=Q(1/a-1/b)/(4\pi\epsilon_0)$. Thus its <capacitance> and <electrostatic energy> are
$$
\boxed{C=\frac{Q}{\Delta\phi}=\frac{4\pi\epsilon_0ab}{b-a},\qquad W=\frac{Q^2}{2C}.}
$$
The energy follows either by charging work $\int_0^Q q/C\,dq$, or directly from $W=(\epsilon_0/2)\int|\mathbf E|^2\,d^3x=Q^2(1/a-1/b)/(8\pi\epsilon_0)$.
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