#### Answer

$P_B = P_C \gt P_A$

#### Work Step by Step

We can find $P_A$:
$P_A = \frac{E_A}{t_A}$
$P_A = \frac{2~J}{2~s}$
$P_A = 1~J/s$
$P_A = 1~W$
We can find $P_B$:
$P_B = \frac{E_B}{t_B}$
$P_B = \frac{10~J}{5~s}$
$P_B = 2~J/s$
$P_B = 2~W$
We can find $P_C$:
$P_C = \frac{E_C}{t_C}$
$P_C = \frac{2\times 10^{-3}~J}{1\times 10^{-3}~s}$
$P_C = 2~J/s$
$P_C = 2~W$
We can rank the power of the sound waves in order from largest to smallest.
$P_B = P_C \gt P_A$