= Solution
Put
$$
k=\frac{\sqrt{2mE}}{\hbar},
\qquad
K=\frac{\sqrt{2m(E+V_0)}}{\hbar}.
$$
The <Time-independent Schrödinger equation> has the forms
$$
\psi(x)=
\begin{cases}
Ae^{ikx}+Be^{-ikx},&x<0,\\
Ce^{iKx}+De^{-iKx},&0\leq x\leq a,\\
Fe^{ikx},&x>a.
\end{cases}
$$
The first region contains the incident and <reflected wave>[reflected waves], while the final region contains only the transmitted wave.
Continuity of $\psi$ and $\psi'$ at both edges of the <finite square well> gives the standard transmission coefficient
$$
T=\frac{|F|^2}{|A|^2}
=\left[
1+\frac{(K^2-k^2)^2}{4k^2K^2}\sin^2(Ka)
\right]^{-1}.
$$
When $V_0=3E$, one has $K=2k$, and therefore
$$
\frac{(K^2-k^2)^2}{4k^2K^2}=\frac9{16},
\qquad
Ka=\frac{a\sqrt{8mE}}{\hbar}.
$$
Hence the <transmission probability> is
$$
\boxed{
T=\frac{16}{16+9\sin^2\!\left(a\sqrt{8mE}/\hbar\right)}
}.
$$
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