Answer
$6.3\times10^{-7}F$
Work Step by Step
Let $Q_i$ and $V_i$ denote the initial charge and voltage on the capacitor, and let $Q_f$ and $V_f$ denote the final charge and voltage.
Equation 17–7, Q=CV, relates the charges and voltages to the capacitance.
$Q_i=CV_i$ and $Q_f=CV_f$
$$Q_f-Q_i=CV_f-CV_i= C(V_f-V_i)$$
$$C=\frac{ Q_f-Q_i }{ V_f-V_i }=\frac{15\times10^{-6}C}{24V}=6.3\times10^{-7}F$$
