Chapter 20 - Electric Potential and Electric Potential Energy - Problems and Conceptual Exercises - Page 720: 64

$C=0.000227F$

To find the capacitance, use the formula for potential energy of a capacitor $$U=\frac{1}{2}CV^2$$ Solving for capacitance $C$ yields $$C=\frac{2U}{V^2}$$ Substituting known values of $U=125J$ and $V=1050V$ yields a capacitance of $$C=\frac{2(125J)}{(1050V)^2}=0.000227F$$