Answer
$\sqrt{\frac{3}{2}}$.
Work Step by Step
Outside the well, for $x \gt L$, $U=U_o$ and the momentum $p=\sqrt{2m(E-U_o)}= \sqrt{2m(2U_o)}$.
Inside the well, for $0 \lt x \lt L$, $U=0$ and the momentum $p=\sqrt{2m(E-0)}= \sqrt{2m(3U_o)}$.
The de Broglie wavelength is inversely proportional to the momentum, $\lambda=\frac{h}{p}$.
The ratio of wavelengths is now calculated.
$$\frac{\lambda_{out}}{\lambda_{in}}=\frac{\sqrt{2m(3U_o)}}{ \sqrt{2m(2U_o)}}$$
$$\frac{\lambda_{out}}{\lambda_{in}}=\sqrt{\frac{3}{2}}$$
