Answer
4.2 million volts.
Work Step by Step
The kinetic energy that the proton has after being accelerated, which is $qV_{initial}$, is changed to potential energy just as the proton’s outer edge reaches the outer radius of the silicon nucleus, because its kinetic energy is zero at that instant.
$$KE_i=PE_f$$
$$eV_{initial}=k\frac{e(14e)}{r}$$
$$V_{initial}=k\frac{14e}{r}$$
$$V_{initial}=(8.988\times10^9Nm^2/C^2)\frac{14(1.60\times10^{-19}C)}{(1.2+3.6)\times10^{-15}m}$$
$$V_{initial}=4.2\times10^6\;volts$$
