Solving for the energies of this system analytically does involve solving a trancendental equation numerically, if memory serves. There's nothing wrong with that, but it can be a bit difficult to clearly see the influences of the various parameters on the result.
A different approach is to treat this problem with perturbation theory. Since you're assuming that the step height is small$^\dagger$, a good start would be to calculate the first order corrections to the energy eigenvalues.
Explicitly, let your Hamiltonian be
$$\hat H = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+ \lambda V(x), \qquad V(x)=\cases{1 & $x\in \left[\frac{L}{2}-\frac{a}{2},\frac{L}{2}+\frac{a}{2}\right]$\\0 & else}$$
This is the Hamiltonian for an infinite potential well with a potential step of width $a$ and height $\lambda$ in the center. To first order in $\lambda$, the corrected energies are simply
$$E_n \simeq E_n^{(0)}+ \lambda \left<\psi_n^{(0)}|\hat V |\psi_n^{(0)}\right> = E_n^{(0)} + \lambda \int_{L/2-a/2}^{L/2+a/2}\psi_n^{(0)*}\psi_n^{(0)} dx$$
where $E_n^{(0)}$ and $\psi_n^{(0)}$ are the uncorrected energies and (normalized) eigenvectors, respectively. We already know what those are from the elementary solution of the infinite potential well, so by evaluating that integral you can see how those energies will change when you introduce the step - at least as long as the step height is small.
$^\dagger$ What it means for an operator to be small can be a subtle issue. In this case, we'd want that $\lambda$ be much smaller than the expected value of the unperturbed Hamiltonian in any state of interest. In this case, that would be accomplished if
$$\lambda \ll \frac{\pi^2\hbar^2}{2mL^2}$$
If $\lambda$ exceeds this limit, then the first order correction will no longer be a good approximation of how the energy will have changed.
