Answer
$$N=4.44 \times 10^{5} \mathrm{rev}$$
$$d \approx 1.2 \times 10^{10} \mathrm{m}$$
$$t=4.0 \times 10^{1} \mathrm{s}$$
Work Step by Step
$\text{These protons will be moving at essentially speed of the light for entire time of}$ $\text{acceleration.}$
$\text{The number of revolutions is }$
$$\frac{\mathrm {the\ total\ gain\ in\ energy} }{\mathrm{the\ energy\ gain\ per\ revolution}}$$
$\text{Then the distance is }$
$\text{the number of revolutions} \times \text{the circumference of the ring}$
$\text{, and the time is }$
$$\frac{\mathrm {the\ distance\ of\ travel\ } }{\mathrm{the\ speed\ of\ the\ protons}}$$
$Then$
$$N=\frac{\Delta E}{\Delta E / \mathrm{rev}}=\frac{\left(4.0 \times 10^{12} \mathrm{eV}-450 \times 10^{9} \mathrm{eV}\right)}{8.0 \times 10^{6} \mathrm{eV} / \mathrm{rev}}=4.44 \times 10^{5} \mathrm{rev}$$
$$d=N(2 \pi R)=\left(4.44 \times 10^{5}\right) 2 \pi\left(4.3 \times 10^{3} \mathrm{m}\right)=$$$$1.199 \times 10^{10} \mathrm{m} \approx 1.2 \times 10^{10} \mathrm{m}$$
$$t=\frac{d}{c}=\frac{1.199 \times 10^{10} \mathrm{m}}{3.00 \times 10^{8} \mathrm{m} / \mathrm{s}}=$$
$$=4.0 \times 10^{1} \mathrm{s}$$
