Solution:

The electric field \( E \) is related to the potential difference \( \Delta V \) by:

\[
\Delta V = -\int_{x_1}^{x_2} E \, dx
\]

Given \( E = \frac{100}{x^2} \), the potential difference between \( x = 10 \) m and \( x = 20 \) m is:

\[
\Delta V = -\int_{10}^{20} \frac{100}{x^2} \, dx
\]

\[
\Delta V = -\left[ \frac{100}{x} \right]_{10}^{20}
\]

\[
\Delta V = -\left( \frac{100}{20} - \frac{100}{10} \right)
\]

\[
\Delta V = -\left( 5 - 10 \right) = 5 \, \text{V}
\]

Thus, the potential difference is \( 5 \, \text{V} \).