 Solution

The infinite exterior potential enforces $\psi =0$ outside $0<x<L$. Within the well,The boundary conditions give $B=0$ and $kL=n\pi$, with $n\ge 1$. Since ${\int }_0^L{\sin }^2(n\pi x/L)dx=L/2$, normalization givesThese functions vanish outside the well. The ground state $n=1$ has energy $E_1$ and one density maximum at $L/2$. The first excited state $n=2$ has energy $4E_1$, a central node, and maxima at $L/4,3L/4$. The second excited state $n=3$ has energy $9E_1$, nodes at $L/3,2L/3$, and maxima at $L/6,L/2,5L/6$.

There is no normalized $n=0$ state: that choice gives the zero function. The original wells figure shows all three wavefunctions and probability densities. An arbitrary overall wavefunction sign does not affect the probability distribution.