We first find the normalization constant $A$ , which we can take to be real and positive.
$\begin{align*}\int_0^L \left|\psi\right|^2dx &= A^2\int_0^L\sin^2\left(\frac{3 \pi x}{L} \right)dx \\&= A^2 \int_0^L ...$
So, in order for $\psi$ to be normalized, $A$ must equal $\sqrt{2/L}$ . So,the desired probability is
$\begin{align*}\int_{L/3}^{2L/3} \left|\psi\right|^2dx &= A^2\int_{L/3}^{2L/3}\sin^2\left(\frac{3 \pi x}{L} \right)dx \\&a...$
Therefore, answer (B) is correct.

Solution to 2008 Problem 85