= Solution
The <wave numbers> before and after the drop are $k=\sqrt{2mE}/\hbar$ and $q=\sqrt{2m(3E)}/\hbar=\sqrt3\,k$. Continuity of $\psi$ and $\psi'$ at the finite step gives
$$
1+r=t,\qquad k(1-r)=qt,\qquad r=\frac{k-q}{k+q}.
$$
Thus the reflected probability flux fraction is
$$
\boxed{R=|r|^2=\left(\frac{1-\sqrt3}{1+\sqrt3}\right)^2=7-4\sqrt3\simeq0.0718.}
$$
The transmission probability is $(q/k)|t|^2=1-R$, rather than simply $|t|^2$. Reflection is possible at a downward <potential step> even though both sides are classically allowed.
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