There's a lot you can do, but the short of it is that the eigen-energy quantization condition still boils down to a transcendental equation that must be solved numerically.
Start with the wavefunction Ansatz
$$
\psi(x) = \begin{cases}
A \sin(kx) & 0<x<a \\
B \sin(k(L-x)) & a<x<L
\end{cases}
$$
where $E = \hbar^2 k^2/2m$.
To find the effect of the delta-function potential, start with the Schrödinger equation,
$$
\left[-\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + V_0\delta(x-a)\right] \psi(x) = E\psi(x),
$$
and integrate it between $a-\epsilon$ and $a+\epsilon$ to get, in the $\epsilon\to0$ limit,
$$
-\frac{\hbar^2}{2m}(\psi'(a^+)-\psi'(a^-)) + V_0 \psi(a) = 0.
$$
When putting in our Ansatz, this reads
$$
-\frac{\hbar^2}{2m}(-Bk\cos(k(L-a))-Ak\cos(ka)) + V_0 \frac12 (A\sin(ka) + B\sin(k(L-a)) = 0,
$$
which is further supplemented by the continuity equation
$$
A\sin(ka) = B\sin(k(L-a))
$$
to produce the coupled set of equations
\begin{align}
A\sin(ka) - B\sin(k(L-a)) &= 0 \\
\left[\frac{\hbar^2}{2m}k\cos(ka) + \frac12V_0\sin(ka)\right]A +
\qquad\qquad\qquad\qquad\qquad\qquad & \\
+\left[\frac{\hbar^2}{2m}k\cos(k(L-a))+\frac12V_0\sin(k(L-a))\right] B &= 0,
\end{align}
or a bit more simply
\begin{align}
A\sin(ka) - B\sin(k(L-a)) &= 0 \\
\left[\frac{\hbar^2}{2m}k\cos(ka) + V_0\sin(ka)\right]A
+\frac{\hbar^2}{2m}k\cos(k(L-a)) B &= 0,
\end{align}
which are best expressed in matrix form, as
$$
\begin{pmatrix}
\sin(ka) & - \sin(k(L-a)) \\
\frac{\hbar^2}{2m}k\cos(ka) +V_0\sin(ka) &
\frac{\hbar^2}{2m}k\cos(k(L-a))
\end{pmatrix}
\begin{pmatrix}A \\ B\end{pmatrix}
=
\begin{pmatrix} 0\\0 \end{pmatrix}.
$$
Since we're looking for a nonzero solution, we require this system to be singular, which means that we require the determinant to vanish,
$$
\det\mathopen{}
\begin{pmatrix}
\sin(ka) & - \sin(k(L-a)) \\
\frac{\hbar^2}{2m}k\cos(ka) +V_0\sin(ka) &
\frac{\hbar^2}{2m}k\cos(k(L-a))
\end{pmatrix}
\mathclose{}
=
0,
$$
and this is what forms the quantization condition.
To see its explicit form, we start by unpacking the determinant:
$$
\sin(ka)\frac{\hbar^2}{2m}k\cos(k(L-a))+ \sin(k(L-a))\left(\frac{\hbar^2}{2m}k\cos(ka) +V_0\sin(ka)\right) = 0,
$$
or a bit more cleanly, in terms of the characteristic wavevector $\kappa = 2mV_0/\hbar^2$,
$$
k\sin(ka)\cos(k(L-a))+ \sin(k(L-a))\left(k\cos(ka) +\kappa\sin(ka)\right) = 0,
$$
and it seems that this is about as far as you can go in simplifying this quantization condition.
You can make a little bit of extra headway if you assume that the delta-function potential is in the middle of the well, so $a=L-a = L/2$. In this case you have
$$
k\sin(kL/2)\cos(kL/2)+ \sin(kL/2)\left(k\cos(kL/2) +\kappa\sin(kL/2)\right) = 0,
$$
so things factorize,
$$
\sin(kL/2) \left[ 2 k\cos(kL/2) +\kappa\sin(kL/2)\right] = 0.
$$
This means that you either have the fully unperturbed
$$
\sin(kL/2) = 0
$$
(which is natural - the eigenfunctions of the original problem where $kL/\pi$ is even will have a zero at the center, so they don't see the delta-function potential), or you have
$$
2 k\cos(kL/2) +\kappa\sin(kL/2) = 0,
$$
which simplifies to
$$
\cot(kL/2) = -\frac{\kappa}{2k},
$$
but this is still not quite within exactly-solvable territory. This can be easily solved numerically, but it's still outside of the range of what you can deal with using analytical methods.
