Answer
${\bf 1.02 }\; \text{fm}$
Work Step by Step
We know that the penetration distance is given by
$$
\eta = \frac{\hbar}{\sqrt{2m(U_0 - E)}}
$$
We also know that $ \hbar = 1.05 \times 10^{-34} \; \text{J} \cdot \text{s} $, and the mass of the neutron is $ m = 1.67 \times 10^{-27} \; \text{kg} $.
Plug into the formula above.
$$ \eta = \frac{1.05 \times 10^{-34}}{\sqrt{2(1.67 \times 10^{-27})(U_0 - E)}}\tag 1 $$
From Figure 40.17 we can say that: the neutron is trapped in a finite square potential well, and the well's potential depth is $ 50 \; \text{MeV} $, but our neutron is only $ 20 \; \text{MeV} $ below the top of the potential well.
Thus,
$$ U_0 - E = 0-(-20)=20 \; \text{MeV} \text{J} $$
Plug into (1);
$$ \eta = \frac{1.05 \times 10^{-34}}{\sqrt{2(1.67 \times 10^{-27})(20\times 1.6\times 10^{-19} )}} =\bf 1.02 \times 10^{-12}\;\rm m$$
$$
\eta =\color{red}{\bf 1.02 }\; \text{fm}
$$
