Answer
${\bf 0.383}\;\rm nm$
Work Step by Step
We are given that the electron's energy is $ 1.50 \; \rm{eV} $ in a potential well of depth $ 2.00 \; \rm{eV} $, and we need to find the distance into the classically forbidden region where the amplitude of the wave function decreases to 25% of its value at the edge of the well.
We know that the wave function decaying is given by
$$
\psi(x) = \psi_{\rm edge} \;e^{-(x-L)/\eta}
$$
So when $ \psi(x) = 0.25 \psi_{\rm edge}$,
$$
0.25 \color{red}{\bf\not} \psi_{\rm edge}= \color{red}{\bf\not} \psi_{\rm edge} \;e^{-(x-L)/\eta}
$$
Hence,
$$
0.25= \;e^{-(x-L)/\eta}
$$
Taking the natural logarithm of both sides:
$$
\ln(0.25) = -\frac{x-L}{\eta}
$$
Solve for $ x-L $:
$$
x-L = -\eta \ln(0.25)\tag 1
$$
Now we need to find $\eta$;
$$
\eta = \frac{\hbar}{\sqrt{2m(U_0 - E)}}
$$
Plug into (1);
$$
x-L = -\ln(0.25) \; \dfrac{\hbar}{\sqrt{2m(U_0 - E)}}
$$
Plug the given with converting eV to J.
$$
x-L = -\ln(0.25) \; \dfrac{(1.055 \times 10^{-34})}{\sqrt{2(9.11\times 10^{-31})(2- 1.5)\times 1.6\times 10^{-19}}}
$$
$$
x-L=\color{red}{\bf 0.383}\;\rm nm
$$
