Understand the resonance conditions for a closed-end tube.

In a resonance tube (which is typically a closed-end tube due to the water column), resonance occurs when the length of the air column ($L$) is an odd multiple of a quarter wavelength. The formula for resonance lengths is:

$$L_n = \frac{(2n-1)\lambda}{4}$$

where $n = 1, 2, 3, \dots$ represents the order of resonance.

• For the first resonance ($n=1$): $L_1 = \frac{(2(1)-1)\lambda}{4} = \frac{\lambda}{4}$
• For the second resonance ($n=2$): $L_2 = \frac{(2(2)-1)\lambda}{4} = \frac{3\lambda}{4}$
• For the third resonance ($n=3$): $L_3 = \frac{(2(3)-1)\lambda}{4} = \frac{5\lambda}{4}$

The problem states that the first position of resonance is $16.50 \, \text{cm}$ from the open end. So, $L_1 = 16.50 \, \text{cm}$.
Using the formula for the first resonance:

$$L_1 = \frac{\lambda}{4}$$ $$16.50 \, \text{cm} = \frac{\lambda}{4}$$

Now, solve for $\lambda$:

$$\lambda = 4 \times 16.50 \, \text{cm}$$ $$\lambda = 66.00 \, \text{cm}$$

The "next position where resonance occurs" refers to the second resonance ($L_2$).
Using the formula for the second resonance:

$$L_2 = \frac{3\lambda}{4}$$

Substitute the calculated value of $\lambda$:

$$L_2 = \frac{3 \times 66.00 \, \text{cm}}{4}$$ $$L_2 = 3 \times \left(\frac{66.00}{4}\right) \, \text{cm}$$ $$L_2 = 3 \times 16.50 \, \text{cm}$$ $$L_2 = 49.50 \, \text{cm}$$

The distance from the open end to the next position where resonance occurs is $\boxed{\text{49.50 cm}}$.