Answer
See the detailed answer.
Work Step by Step
a) We are given the potential energy of two nucleons interacting via the strong force
$$U=U_0\left[1-e^{-x/x_0}\right]$$
Plug the known;
$$U=\left(6.0\times 10^{-11}\right)\left[1-e^{-x/\left(2.0\times 10^{-15}\right)}\right]$$
See the graph below.
_______________________________________
b) When $x=5\times10^{-15}$ m, the potential energy is given by
$$U=\left(6.0\times 10^{-11}\right)\left[1-e^{-(5\times 10^{-15})/\left(2.0\times 10^{-15}\right)}\right]$$
$$U=E_{total}=\bf 55.1\times 10^{-12}\;\rm J\tag 1$$
which is the total energy as the author told us.
We draw it as a horizontal line in the figure below (the red line).
_______________________________________
c) We know the total energy of the two p[rotons as we found it in part b above. Now we know that when both neutrons collide, some of this total energy will be converted to kinetic energy plus the potential energy between them when the distance between them is twice the radius of the neutron.
Thus,
$$E_{total}=E_f$$
$$E_{total}=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2+U_f$$
The two neutrons are having the same mass and the same final speed since they are identical.
$$E_{total}= m v ^2+ \left(6.0\times 10^{-11}\right)\left[1-e^{-(2R)/\left(2.0\times 10^{-15}\right)}\right]$$
Plugging from (1);
$$55.1\times 10^{-12}= m v ^2+ \left(6.0\times 10^{-11}\right)\left[1-e^{-(2R)/\left(2.0\times 10^{-15}\right)}\right]$$
Thus,
$$mv^2=(55.1\times 10^{-12})- \left(6.0\times 10^{-11}\right)\left[1-e^{-(2R)/\left(2.0\times 10^{-15}\right)}\right]$$
$$ v =\sqrt{\dfrac{(55.1\times 10^{-12})- \left(6.0\times 10^{-11}\right)\left[1-e^{-(2R)/\left(2.0\times 10^{-15}\right)}\right]}{m}}$$
Plugging the known;
$$ v =\sqrt{\dfrac{(55.1\times 10^{-12})- \left(6.0\times 10^{-11}\right)\left[1-e^{-(1\times 10^{-15})/\left(2.0\times 10^{-15}\right)}\right]}{1.67\times 10^{-27}}}$$
$$v=\color{red}{\bf 1.37\times 10^{8}}\;\rm m/s$$
