Answer
The ratio of of the intensity of the wave in water to that of the wave in air is $~~2.81\times 10^{-4}$
Work Step by Step
Note that $p_m = v~\rho~\omega~s_m$
We can find an expression for the pressure amplitude of a sound wave in the water:
$I_w = \frac{1}{2}~\rho_w~v_w~\omega^2~s_w^2$
$I_w = \frac{1}{2}~\rho_w~v_w~\omega^2~(\frac{p_w}{v_w~\rho_w~\omega})^2$
$I_w = \frac{p_w^2}{2~v_w~\rho_w}$
$p_w^2 = I_w~2~v_w~\rho_w$
$p_w = \sqrt{I_w~2~v_w~\rho_w}$
We can find an expression for the pressure amplitude of a sound wave in the air:
$I_a = \frac{1}{2}~\rho_a~v_a~\omega^2~s_a^2$
$I_a = \frac{1}{2}~\rho_a~v_a~\omega^2~(\frac{p_a}{v_a~\rho_a~\omega})^2$
$I_a = \frac{p_a^2}{2~v_a~\rho_a}$
$p_a^2 = I_a~2~v_a~\rho_a$
$p_a = \sqrt{I_a~2~v_a~\rho_a}$
We can find the ratio $\frac{I_w}{I_a}$:
$p_w = p_a$
$\sqrt{I_w~2~v_w~\rho_w} = \sqrt{I_a~2~v_a~\rho_a}$
$I_w~2~v_w~\rho_w = I_a~2~v_a~\rho_a$
$\frac{I_w}{I_a} = \frac{v_a~\rho_a}{v_w~\rho_w}$
$\frac{I_w}{I_a} = \frac{(343~m/s)(1.21~kg/m^3)}{(1482~m/s)(998~kg/m^3)}$
$\frac{I_w}{I_a} = 2.81\times 10^{-4}$
The ratio of of the intensity of the wave in water to that of the wave in air is $~~2.81\times 10^{-4}$
