The problem describes a resonance tube closed at one end. For a tube closed at one end, the resonant frequencies are given by the formula:

The problem describes a resonance tube closed at one end. For a tube closed at one end, the resonant frequencies are given by the formula:

$$f_n = \frac{(2n-1)v}{4L}$$

where $v$ is the speed of sound, $L$ is the length of the tube, and $n = 1, 2, 3, \dots$ represents the harmonic number (odd harmonics only).

The lowest frequency corresponds to $n=1$. This is the fundamental frequency or the first harmonic.

$$f_1 = \frac{(2(1)-1)v}{4L} = \frac{1v}{4L} = \frac{v}{4L}$$

The next resonant frequency after the fundamental corresponds to $n=2$. This is the first overtone or the third harmonic.

$$f_2 = \frac{(2(2)-1)v}{4L} = \frac{3v}{4L}$$

Now, we find the ratio of the next resonant frequency to the lowest resonant frequency:

$$\frac{f_2}{f_1} = \frac{\frac{3v}{4L}}{\frac{v}{4L}}$$ $$\frac{f_2}{f_1} = \frac{3v}{4L} \times \frac{4L}{v}$$ $$\frac{f_2}{f_1} = 3$$

The ratio $\frac{f_2}{f_1}$ is 3.

The final answer is $\text{B. } 3$.