Answer
$s = 2.3 \,\text{cm}$
Work Step by Step
The change in potential is equal
\begin{gather*}
\Delta V = \frac{ \lambda}{2 \pi \epsilon_{o}} \ln \frac{r_{f}}{r_{i}}\\
\frac{ \lambda}{2 \pi \epsilon_{o}} \ln \frac{r_{f}}{r_{i}} = 175 \,\text{V}\\
\ln \frac{r_{f}}{r_{i}} =175 \,\text{V} \, \dfrac{ 2 \pi ( 8.85 \times 10^{-12}\,\text{F/m})}{15 \times 10^{-9}\,\text{C/m}}\\
\ln \frac{r_{f}}{r_{i}} = 0.65 \\
r_{f}=r_{i} e^{ 0.65} \\
r_{f}=0.025 \,\text{m} \,e^{ 0.65}\\
r_f = 0.048 \,\text{m}
\end{gather*}
The distance between the cylinder and the other probe is
$$s = r_f -R = 0.048 \,\text{m} - 0.025 \,\text{m} = 0.023 \,\text{m} =\boxed{ 2.3 \,\text{cm}}$$
