Answer
$2\Big(\frac{h}{8mL^2}\Big)$
Work Step by Step
The frequency $(v)$ that corresponds to photon energy $(E)$ is given by
$v=\frac{E}{h}$
From the section $(a)$ of this problem, we can say that the second lowest frequency occurs at $E_{3,1,1}\leftrightarrow E_{2,2,1} $ transion. The correspond frequency is
$v=\frac{\Delta E}{h}=2\Big(\frac{h}{8mL^2}\Big)$
