 Solution

The grounded outer cylinder fixes $\Phi (a)=0$. Integrating the annular electric field inward givesEquivalently, the coaxial cylindrical capacitor has $C_{\mathrm{cyl}}=2\pi {ϵ}_{0}l/\log (a/b)$ and carries charge $Q=4\pi {ϵ}_{0}RV$, so $\Phi (b)=Q/C_{\mathrm{cyl}}$. The original sphere potential $V$ is used to infer its conserved charge; it is not the final potential of the sphere or of the inner cylinder.