= Solution
Write $E=-\hbar^2\kappa^2/(2m)$. An even <bound state> is proportional to $\cos(kx)$ inside the well and to $e^{-\kappa|x|}$ outside, where
$$
k^2=\frac{m}{a\hbar^2}-\kappa^2.
$$
Continuity of the logarithmic <derivative> at $a$ gives
$$
\kappa=k\tan(ka).
$$
For sufficiently small $a$, $ka$ lies in the first monotone branch, so this equation has exactly one positive root and the even state is unique up to scale.
As $a\downarrow0$, $\kappa\sim k^2a\to m/\hbar^2$. Hence
$$
E_0=-\frac{m}{2\hbar^2}.
$$
The wells converge distributionally to $V_0(x)=-\delta(x)$, whose normalized even <bound state> is
$$
\psi_0(x)=\sqrt{\frac m{\hbar^2}}e^{-m|x|/\hbar^2}.
$$
Its <derivative> jump is precisely the one imposed by the <delta potential>.
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