= Solution
Matching $\psi$ and $\psi'$ at both walls, or propagating the pair $(\psi,\psi')$ through the interior by the matrix $\bigl(\begin{smallmatrix}\cos\zeta a&\sin\zeta a/\zeta\\-\zeta\sin\zeta a&\cos\zeta a\end{smallmatrix}\bigr)$, gives
$$
t(k)=\frac{e^{-ika}}{D(k)},\qquad D(k)=\cos(\zeta a)-i\frac{k^2+\zeta^2}{2k\zeta}\sin(\zeta a).
$$
Here $\zeta^2=k^2+2mU_0/\hbar^2$. For real positive $k$, taking the modulus gives
$$
\boxed{T=\left[\cos^2\zeta a+\frac{(k^2+\zeta^2)^2}{4k^2\zeta^2}\sin^2\zeta a\right]^{-1}.}
$$
The equal exterior velocities make this also the current <transmission coefficient>.
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