The pole is vertically submerged in the swimming pool, and the length of the shadow is due to the sunlight that is incident at an angle of $30^{\circ}$ with the surface of water. Hence, the angle of incidence, $i$, is $60^{\circ}$ (since the angle of incidence is measured from the normal to the surface, and the normal is perpendicular to the surface).

Using Snell's law, we find the angle of refraction, $r$, as you correctly did:

$$\sin r = \frac{n_{1}}{n_{2}} \sin i = \frac{3}{4} \sin 60^{\circ} = \frac{3 \sqrt{3}}{8}$$

This gives $\tan r = \frac{3 \sqrt{3}}{\sqrt{37}}$.

Now, the shadow of the pole in the water forms a right triangle, with the submerged part of the pole as one side, the shadow as the hypotenuse, and the line segment from the water surface to the end of the shadow as the other side. The angle at the water surface is $r$. Thus, we can write the following relationships:

The length of the submerged part of the pole (which I'll call $x$), is given by:

$$x = 2.15 \, \text{m} \cdot \cos r$$

And the length of the line segment from the water surface to the end of the shadow (which I'll call $y$), is given by:

$$y = 2.15 \, \text{m} \cdot \sin r$$

The total length of the pole above the water surface is then the sum of $x$ and the depth of the swimming pool (1.5 m), which gives us the equation:

$$x \sqrt{3} + 1.5 \, \text{m} \cdot \tan r = 2.15 \, \text{m}$$

Solving this for $x$ gives us:

$$x = \frac{2.15 \, \text{m}}{\sqrt{3}} - \frac{1.5 \, \text{m} \cdot 3}{\sqrt{37}} = 0.502 \, \text{m} = 50.2 \, \text{cm}$$

So, the height of the pole above the water surface is approximately 50 cm.