Answer
The required speed of the detector is $~~\frac{2}{3}~v$
Work Step by Step
We can find an expression for frequency of the third harmonic in tube 2:
$f = \frac{nv}{2L},~~$ where $n = 1,2,3,...$
$f = \frac{3v}{2L}$
Note that the fundamental frequency is $\frac{1}{3}$ of the third harmonic frequency.
We can find the required speed of the detector:
$f' = f~\frac{v-v_D}{v}$
$\frac{f'}{f} = \frac{v-v_D}{v}$
$\frac{1}{3} = \frac{v-v_D}{v}$
$\frac{1}{3}~v = v-v_D$
$v_D = v-\frac{1}{3}~v$
$v_D = \frac{2}{3}~v$
The required speed of the detector is $~~\frac{2}{3}~v$
