Answer
77 microfarads.
Work Step by Step
After the first capacitor is disconnected from the battery, we know the charge on it.
$$Q_{1i}=C_1V_i$$
After this is connected to the second capacitor, the voltage across each capacitor is the same.
We can find the charge on each one.
We know that $Q_{1f}=C_1V_f$ and $Q_{2f}=C_2V_f$.
Charge is conserved.
$$C_1V_i= C_1V_f+ C_2V_f$$
$$C_2=C_1(\frac{V_i}{V_f}-1)= (7.7\times10^{-6} F)(\frac{165V}{15V}-1)= 7.7\times10^{-5} F $$
