1. Energy gained by proton in a cyclotron

In a cyclotron, each time the proton crosses the gap, it gains energy $\Delta K = eV$ where $V$ is the accelerating potential.

Given maximum accelerating potential: $V = 12\,\text{kV} = 12 \times 10^3\,\text{V}$

So energy gained in one gap crossing is $\Delta K = eV = 1.6\times 10^{-19} \times 12\times 10^3$ $\Delta K = 19.2\times 10^{-16} = 1.92\times 10^{-15}\,\text{J}$

2. Energy gained in one complete revolution

In one full revolution, the proton crosses the gap twice, so energy gained per revolution is $\Delta K_{\text{rev}} = 2eV$ $\Delta K_{\text{rev}} = 2 \times 1.92\times 10^{-15}$ $\Delta K_{\text{rev}} = 3.84\times 10^{-15}\,\text{J}$

3. Final kinetic energy required

Required speed: $v = \frac{c}{6} = \frac{3\times 10^8}{6} = 5\times 10^7\,\text{m/s}$

Since this speed is much less than $c$, non-relativistic formula is acceptable: $K = \frac{1}{2}mv^2$

Thus, $K = \frac{1}{2} \times 1.67\times 10^{-27} \times (5\times 10^7)^2$

Now, $(5\times 10^7)^2 = 25\times 10^{14} = 2.5\times 10^{15}$

So, $K = \frac{1}{2} \times 1.67\times 10^{-27} \times 2.5\times 10^{15}$ $K = 0.5 \times 4.175\times 10^{-12}$ $K = 2.0875\times 10^{-12}\,\text{J}$

4. Number of revolutions

If $N$ is the number of revolutions, then $N(2eV) = K$

Hence, $N = \frac{K}{2eV} = \frac{2.0875\times 10^{-12}}{3.84\times 10^{-15}}$

$N \approx 543.6$

Since the number of complete revolutions is taken as the nearest integer, $N \approx 543$

5. Final answer

The number of revolutions made by the proton is $\boxed{543}$