Answer
According to Table 25-1, we should use Pyrex.
Work Step by Step
We can find the required capacitance:
$\frac{1}{2}CV^2 = U$
$C = \frac{2U}{V^2}$
$C = \frac{(2)(7.4\times 10^{-6}~J)}{(652~V)^2}$
$C = 34.815\times 10^{-12}~F$
We can find the required dielectric constant of the material:
$\kappa = \frac{34.815\times 10^{-12}~F}{7.4~\times 10^{-12}~F} = 4.7$
According to Table 25-1, we should use Pyrex.
