Answer
2.8 nC.
Work Step by Step
Use equation 17–5, which gives the potential V of a point charge.
$$V=165volts=\frac{1}{4\pi\epsilon_o}\frac{Q}{r}$$
$$Q=4\pi\epsilon_orV$$
$$=(\frac{C^2}{8.99\times10^9 N\cdot m^2})(0.15m)(165V)$$
$$=2.8\times10^{-9}C$$