Answer
(a) $\Delta V = 78 \times 10^3 \mathrm{V}$
(b) $\Delta V = 0$
Work Step by Step
(a) The potential in the cylinder is given by
\begin{align}
\Delta V &= \frac{\lambda}{2 \pi \epsilon_{o}} \ln \frac{r}{R}\\
&= \frac{8.5 \times 10^{-6}}{2 \pi \times 8.85 \times 10^{-12}} \ln \left(\frac{10}{6}\right)\\
&= \boxed{78 \times 10^3 \mathrm{V}}
\end{align}
(b) At distance 1 cm, the potential inside the cylinder is $\textbf{zero}$.
