Answer
a) $2\times 10^{-12}\;\rm J$
b) $1\times 10^{-12}\;\rm J$
Work Step by Step
$$\color{blue}{\bf [a]}$$
We need to recall the energy problems we solved earlier in the energy chapters, this one is similar to them [chapter 10] but here we use electric potential energy plus the other energies.
The electric force is a conservative force like the gravitational force. So we can say that the energy here is conserved.
Hence,
$$E_i=E_f$$
$$K_i+U_{ei}=K_f+U_{ef}\tag 1$$
The final speed of the proton is zero, and the lead nucleus is stationary and remains stationary.
So the final kinetic energy of the system when the proton is at the turning point is zero.
$$K_i+U_{ei}=0+U_{ef} $$
And since the proton is released from a very far away distance, the initial electric potential energy of the proton is zero.
$$K_i+0= U_{ef}$$
$$K_i = U_{ef}$$
So at 10 fm from the nucleus, the electric potential energy, from the given in the graph, is $2\times 10^{-12}$ J.
Thus,
$$K_i =\color{red}{\bf 2\times 10^{-12}}\;\rm J$$
$$\color{blue}{\bf [b]}$$
Now we know the initial kinetic energy of the proton, and we need to find its kinetic energy at 20 fm. So we can use (1) again; where $U_{ei}=0$
$$K_i+0=K_f+ U_{ef}$$
Thus,
$$K_f =K_i- U_{ef}$$
at 20 fm, $U_e=1\times 10^{-12}$ J
$$K_f =( 2\times 10^{-12})- ( 1\times 10^{-12})$$
$$K_f =\color{red}{\bf 1\times 10^{-12}}\;\rm J$$
