Answer
The ratio of of the pressure amplitude of the wave in water to that of the wave in air is $~~59.7$
Work Step by Step
Note that $p_m = v~\rho~\omega~s_m$
We can find an expression for the intensity of a sound wave in the water:
$I_w = \frac{1}{2}~\rho_w~v_w~\omega^2~s_w^2$
$I_w = \frac{1}{2}~\rho_w~v_w~\omega^2~(\frac{p_w}{v_w~\rho_w~\omega})^2$
$I_w = \frac{p_w^2}{2~v_w~\rho_w}$
We can find an expression for the intensity of a sound wave in the air:
$I_a = \frac{1}{2}~\rho_a~v_a~\omega^2~s_a^2$
$I_a = \frac{1}{2}~\rho_a~v_a~\omega^2~(\frac{p_a}{v_a~\rho_a~\omega})^2$
$I_a = \frac{p_a^2}{2~v_a~\rho_a}$
We can find the ratio $\frac{p_w}{p_a}$:
$I_w = I_a$
$\frac{p_w^2}{2~v_w~\rho_w} = \frac{p_a^2}{2~v_a~\rho_a}$
$\frac{p_w^2}{p_a^2} = \frac{v_w~\rho_w}{v_a~\rho_a}$
$\frac{p_w}{p_a} = \sqrt{\frac{v_w~\rho_w}{v_a~\rho_a}}$
$\frac{p_w}{p_a} = \sqrt{\frac{(1482~m/s)(998~kg/m^3)}{(343~m/s)(1.21~kg/m^3)}}$
$\frac{p_w}{p_a} = 59.7$
The ratio of of the pressure amplitude of the wave in water to that of the wave in air is $~~59.7$
