Step 1: Identify the type of resonance tube and its resonant frequencies. 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). Step 2: Determine the lowest resonant frequency ($f_1$). The lowest frequency corresponds to $n=1$. This is the fundamental frequency or the first harmonic. $$f_1 = \frac{(2(1)-1)v}{4L} = \frac{1v}{4L} = \frac{v}{4L}$$ Step 3: Determine the next resonant frequency ($f_2$). The next resonant frequency after the fundamental corresponds to $n=2$. This is the first overtone or the third harmonic. $$f_2 = \frac{(2(2)-1)v}{4L} = \frac{3v}{4L}$$ Step 4: Calculate the ratio $\frac{f_2}{f_1}$. 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 $\boxed{\text{B. } 3}$.