= Solution
The grounded outer cylinder fixes $\Phi(a)=0$. Integrating the annular <electric field> inward gives
$$
\boxed{\Phi(b)=\int_b^aE_r(r)\,dr=\frac{2RV}{l}\log\frac ab.}
$$
Equivalently, the <coaxial cylindrical capacitor> has $C_{\mathrm{cyl}}=2\pi\epsilon_0l/\log(a/b)$ and carries charge $Q=4\pi\epsilon_0RV$, 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.
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