Answer
The smallest value of x locating a node is $x = 0$
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 $x = L$
The smallest value of x locating a node is $x = 0$