= Solution
Use the <acoustic velocity potential> convention
$$
\mathbf u=\nabla\phi,
\qquad
\widetilde p=-\rho_0\frac{\partial\phi}{\partial t}.
$$
Put
$$
k=\frac{\omega}{c_0},
\qquad
\alpha=kR.
$$
For a spherically symmetric harmonic solution, write
$$
\phi(r,t)=F(r)\sin\omega t.
$$
The wave equation and
$$
\nabla^2F=\frac1r\frac{d^2}{dr^2}(rF)
$$
give the radial Helmholtz equation
$$
\frac{d^2}{dr^2}(rF)+k^2(rF)=0.
$$
The rigid inner sphere requires
$$
u_r(R,t)=F'(R)\sin\omega t=0.
$$
A form satisfying this condition automatically is
$$
F(r)=\frac C r
\left[
\cos k(r-R)+\frac1{kR}\sin k(r-R)
\right].
$$
Indeed, if the bracket is $h(r)$, then $h(R)=1$ and $h'(R)=1/R$, so $(h/r)'=0$ at $R$.
At the outer sphere,
$$
\varepsilon p_0\cos\omega t
=-\rho_0\omega F(2R)\cos\omega t,
$$
so
$$
F(2R)=-\frac{\varepsilon p_0}{\rho_0\omega}.
$$
Since
$$
F(2R)=\frac C{2R}
\left(\cos\alpha+\frac{\sin\alpha}{\alpha}\right),
$$
we obtain
$$
C=-\frac{2R\varepsilon p_0}
{\rho_0\omega(\cos\alpha+\alpha^{-1}\sin\alpha)}.
$$
Therefore
$$
\boxed{
\phi(r,t)
=-\frac{2R\varepsilon p_0}{\rho_0\omega r}
\frac{\cos k(r-R)+(kR)^{-1}\sin k(r-R)}
{\cos\alpha+\alpha^{-1}\sin\alpha}
\sin\omega t}.
$$
The assumption $\alpha\leq\pi/2$ keeps the displayed denominator positive.
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