Answer
5.15 pF.
Work Step by Step
Calculate the voltage across each capacitor.
$$V_{175pF}=\frac{Q}{C}=\frac{125\times10^{-12}C}{175\times10^{-12}F}=0.714V$$
The voltages across the capacitors add to the battery emf, by Kirchhoff’s loop law, so the voltage across the unknown capacitor is 25.0V-0.714V=24.286 volts.
Capacitors in series each have the same amount of charge. The unknown capacitor also stores 125 pC. Finally, calculate the unknown capacitance.
$$C=\frac{Q}{V}=\frac{125\times10^{-12}C}{24.286 V}=5.15\times10^{-12}F $$
