So this is actually a really easy problem. All you do is take the Schrödinger Equation:
[tex]-\dfrac{\hbar^2}{2m}\dfrac{d^2\psi}{dx^2}+V(x)\psi(x) = E\psi(x)[/tex]
Now you plug in the potential. This can be centered at 0, but it's usually more easily centered at L/2, where L is the well width. The potential is,
[tex]V(x) = 0[/tex] when 0<x<L, and
[tex]V(x) = \infty[/tex] when x<0 or x>L
So inside the well you've just got:
[tex]-\dfrac{\hbar^2}{2m}\dfrac{d^2\psi}{dx^2} = E\psi(x)[/tex]
Or,
[tex]\dfrac{d^2\psi}{dx^2}+ \dfrac{2mE}{\hbar^2}\psi(x)=0[/tex]
What does this look like? Hint: it's exactly the same mathematical form as the simple harmonic oscillator from classical mechanics (but it doesn't actually have anything to do with oscillators physics-wise). You should be able to solve the equation using the ansatz,
[tex]\psi(x) = Asin(kx) + Bcos(kx)[/tex]
The value of k is found from the boundary conditions. Remember that the potential outside the well is zero. Thus the wavefunction has to either be 0 or infinity at those places. Common sense dictates it'll be zero, so the wave has to "fit" inside the box. Thus the wavenumber will have to be such that the wave always goes to zero at the boundaries. This eliminates all cosine solutions, since cos(0)=1. You can figure out what k is. From that you can solve for the energy. And if you really want, you can figure out what A is by finding the complex modulus squared of psi and treating it as a probability density.
Let me know what you come up with!