Chapter 21 - Electric Current and Direct-Current Circuits - Problems and Conceptual Exercises - Page 761: 112

$0.26M\Omega$

We can find the required resistance as follows: $\frac{1}{R}=-\frac{C}{t}\ln (1-\frac{V}{\mathcal{E}})$ We plug in the known values to obtain: $\frac{1}{R}=-\frac{110\times 10^{-6}F}{\frac{60s}{75}}\ln(1-\frac{0.25V}{9.0V})$ $\frac{1}{R}=3.87\times 10^{-6}\Omega^{-1}$ $R=0.26\times 10^{-6}\Omega$ $R=0.26M\Omega$