Answer
See answers.
Work Step by Step
a. The time constant is RC.
$$\tau=RC$$
Solve for the capacitance.
$$C=\frac{\tau}{R}=\frac{18.0\times10^{-6}s}{15.0\times10^{3}\Omega}=1.20\times10^{-9}F$$
b. The battery has an emf of 24.0 V. When the voltage across the resistor is 16.0 V, the voltage across the capacitor will be 8.0 V. Use the equation, 19-7a, for the voltage across a charging capacitor.
$$V_C=\epsilon(1-e^{-t/\tau})$$
$$t=-\tau\; ln(1-\frac{V_C}{\epsilon})=-(18.0\times10^{-6}s) ln(1-\frac{8V}{24V})=7.30\times10^{-6}s $$