Answer
The second smallest value of $x$ locating a node is $~~x = 0.40~m$
Work Step by Step
We can draw a representation of the second lowest harmonic frequency in a pipe that is closed at one end:
There is a node at $x = 0$
There is an antinode at $x = \frac{L}{3}$
There is a node at $x = \frac{2L}{3}$
There is an antinode at $x = L$
We can use the information about pipe A to find the frequency:
$f = \frac{nv}{2L},~~$ where $~~n = 3$
$f = \frac{(3)(343~m/s)}{(2)(1.20~m)}$
$f = 428.75~Hz$
We can find the length of Pipe B:
$f = \frac{3v}{4L} = 428.75~Hz$
$L = \frac{3v}{(4)(428.75~Hz)}$
$L = \frac{(3)(343~m/s)}{(4)(428.75~Hz)}$
$L = 0.60~m$
The second smallest value of $x$ locating a node is $\frac{(2)(0.60~m)}{3}$ which is $~~x = 0.40~m$
