Answer
See the detailed answer below.
Work Step by Step
First of all, we need to find the critical angle for the water-air boundary
$$\sin\theta_c=\dfrac{n_{air}}{n_{water}}$$
Hence,
$$\theta_c=\sin^{-1}\left[\dfrac{n_{air}}{n_{water}}\right]=\sin^{-1}\left[\dfrac{1}{1.33}\right]\approx\bf 48.8^\circ$$
$$\color{blue}{\bf [a]}$$
Now we need to find the angles $\theta_1$, $\theta_2$, and $\theta_3$.
If $\theta_3$ is less than $\theta_c$, the ray will penetrate the water-air boundary and be refracted into the air. And if it is greater than $\theta_c$, the ray will be fully reflected into the water.
From the geometry of the figure below, we can see that
$$\theta_1=90^\circ-\phi=90^\circ-\tan^{-1}\left[\dfrac{10}{x} \right]$$
when $x=15$ cm,
$$\theta_1= 90^\circ-\tan^{-1}\left[\dfrac{10}{15} \right]=\bf56.3^\circ$$
By Snell's law, we can see that
$$\theta_2=\sin^{-1}\left[ \dfrac{n_{air}\sin\theta_1}{n_{water}}\right]$$
$$\theta_2=\sin^{-1}\left[ \dfrac{(1)\sin56.3^\circ }{1.33}\right]=\bf 38.7^\circ$$
From the geometry of the figure below, we can see that
$$\theta_3=90^\circ-\theta_2=90^\circ-38.7^\circ$$
$$\theta_3=\color{red}{\bf 51.3^\circ}$$
which is greater than $\theta_c$, so we will get ray 1, not 2.
This means that the ray will experience a full reflection inside the water at an angle of $\theta_3$
$$\color{blue}{\bf [b]}$$
From the geometry of the figure below, we can see that
$$\theta_1=90^\circ-\phi=90^\circ-\tan^{-1}\left[\dfrac{10}{x} \right]$$
when $x=25$ cm,
$$\theta_1= 90^\circ-\tan^{-1}\left[\dfrac{10}{25} \right]=\bf 68.2^\circ$$
By Snell's law, we can see that
$$\theta_2=\sin^{-1}\left[ \dfrac{n_{air}\sin\theta_1}{n_{water}}\right]$$
$$\theta_2=\sin^{-1}\left[ \dfrac{(1)\sin 68.2^\circ }{1.33}\right]=\bf 44.3^\circ$$
From the geometry of the figure below, we can see that
$$\theta_3=90^\circ-\theta_2=90^\circ-44.3^\circ$$
$$\theta_3= {\bf 45.7^\circ}$$
which is less than $\theta_c$, so we will get ray 2, not 1.
This means that the ray will penetrate the boundary and refract into the air.
By Snell's law, we can see that
$$\theta_4=\sin^{-1}\left[ \dfrac{n_{water}\sin\theta_3}{n_{air}}\right]$$
$$\theta_4=\sin^{-1}\left[ \dfrac{(1.33)\sin45.7^\circ}{1}\right]$$
$$\theta_4=\color{red}{\bf 72.2^\circ}$$
$$\color{blue}{\bf [c]}$$
To find the minimum value of $x$, we need to find the angle $\theta_2$ that makes $\theta_3=\theta_c$.
So from the geometry of the figure below,
$$\theta_2=90^\circ-\theta_3=90^\circ-\theta_c=90^\circ-48.8^\circ$$
$$\theta_2=\bf 41.2^\circ$$
By Snell's law, we can see that
$$\theta_1=\sin^{-1}\left[ \dfrac{n_{water}\sin\theta_2}{n_{air}}\right]$$
$$\theta_1=\sin^{-1}\left[ \dfrac{(1.33)\sin41.2^\circ}{1}\right]=\bf 61.2^\circ$$
So from the geometry of the figure below,
$$\phi=90^\circ-\theta_1=90^\circ-61.2^\circ=\bf 28.8^\circ$$
where
$$\tan\phi=\dfrac{10}{x}$$
Hence,
$$x=\dfrac{10}{\tan\phi}=\dfrac{10}{\tan28.8^\circ }=18.2\;\rm cm$$
So the minimum value of $x$ is less than 18.2 cm.
$$x_{minimum}\approx\color{red}{\bf 18}\;\rm cm$$
