The critical angle ${}_c$ is defined as the incident angle beyond which there is no refracted ray. The incident light strikes the boundary and completely reflects. It occurs when the refracted beam has an angle ${}_2=90.$ You can predict the critical angle using Snell's Law:

$n_{1}sin{}_c=n_{2}sin(90)$

Since $sin(90)=1$

$sin{}_c=\frac{n_2}{n_1}$

Solving for ${}_c$ :

${}_c=si{n}^{-1}(\frac{n_2}{n_1})$

Use the last equation to compute the critical angle $($in degrees$)$ when water is on $)$ and air is on the bottom $(n_2=1\mathrm{.}0003).($Don$\text{'}$t forget: make sure your calculator is in DEGREE mode and not radian mode!$).$

Is there a critical angle for total internal reflection when air is on top $(n_1=1\mathrm{.}0003)$ and water is on the bottom $(n_2=1\mathrm{.}33)?$

A$.$ Yes. It does not matter which medium is on top or bottom; the critical angle is exactly the same.

B$.$ Yes and no$.$ You will get total internal reflection, but with a different critical angle if you swap the media.

C$.$ No$.$ You will only get total internal reflection when the incident medium has a higher index than the refractive medium.