Answer
a) $2.3\times 10^9\;\rm K$
b) $153$
Work Step by Step
$$\color{blue}{\bf [a]}$$
We can assume that the energy is conserved. So that
$$K_{i1}+K_{i2}+U_i=K_{f1}+K_{f2}+U_f$$
where $_i\rightarrow$ is just before the collision, and $_f\rightarrow$ is just after it.
Assuming that the two protons will stop after the collision since we are seeking the minimum temperature needed for this reaction.
$$K_{i1}+K_{i2}+U_i=0+0+U_f$$
We can assume that the two protons were far away from each other before the collision where $r\rightarrow\infty$ and hence, $U_i=0$
$$K_{i1}+K_{i2}+0= \dfrac{ke^2}{r_f}$$
Note that $r_f$ is the distance between the centers of the two protons which is equal to the diameter of the proton $D_{\rm p}$.
$$K_{i1}+K_{i2}+0= \dfrac{ke^2}{D_{\rm p}}$$
We are given that the two protons have the same average kinetic energy just before the head-on collision. So that $K_{i1}=K_{i2}=\frac{3}{2}k_BT$
$$2\left[ \frac{3}{2}k_BT\right]= \dfrac{ke^2}{D_{\rm p}}$$
Solving for $T$;
$$T= \dfrac{ke^2}{3D_{\rm p}k_B}$$
Plug the known;
$$T= \dfrac{(9\times 10^9)(1.6\times 10^{-19})^2}{3(2.4\times 10^{-15})(1.38\times 10^{-23}) }$$
$$T=\color{red}{\bf 2.3\times 10^9}\;\rm K$$
$$\color{blue}{\bf [b]}$$
We need to find by what factor each proton’s energy exceeds the average kinetic energy at 15 million K.
$$\dfrac{K_1}{K_2}=\dfrac{\frac{3}{2}k_BT_1}{\frac{3}{2}k_BT_2}=\dfrac{T_1}{T_2}$$
$$\dfrac{K_1}{K_2} =\dfrac{2.3\times 10^9}{15\times 10^6}=\color{red}{\bf 153}$$
This means that the kinetic energy of the protons that undergo the fusion reaction must exceed the proton's average kinetic energy by 153 times.
