= Solution
Write the spatial factors of the stationary <wavefunction> as
$$
\psi(x,t)=e^{-iEt/\hbar}
\begin{cases}
e^{ikx}+R e^{-ikx},&x<-a,\\
C e^{kx}+D e^{-kx},&-a<x<a,\\
T e^{ikx},&x>a,
\end{cases}
$$
where
$$
k=\frac{\sqrt{2mE}}{\hbar}.
$$
Since $U_0-E=E$, the decay constant inside the <potential barrier> is also $k$. Continuity of the <wavefunction> and its first <derivative> at $x=-a$ and $x=a$ gives four linear equations. Solving them yields
$$
T=\frac{e^{-2ika}}{\cosh(2ka)}.
$$
Consequently the transmitted wave is
$$
\boxed{
\psi_{\rm tr}(x,t)
=\frac{\exp\!\left(i[k(x-2a)-Et/\hbar]\right)}
{\cosh(2ka)}
},
\qquad x>a,
$$
and its <probability density> is
$$
\boxed{
\rho_{\rm tr}(x,t)=|\psi_{\rm tr}|^2
=\operatorname{sech}^2(2ka)
}.
$$
This is <finite square barrier transmission at half barrier height>.
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