The formula for the radius (r) of a circle traced out by a charged particle in a uniform magnetic field turns out to be:
r = mv/qB
m = mass of particle
v = speed of particle
q = (electric) charge of particle
B = strength of magnetic field
(we can derive this if you need to, but, for now, I will take it as a given)
If we rearrange this to put v on one side of the equation by itself, it becomes:
v = qBr/m
I take the energy of the particle to mean its kinetic energy (KE)
KE = (1/2)mv2
Putting in the expression for v we got above, for a particle moving in a circle, this becomes:
KE = (1/2)m(qBr/m)2 --> q2B2r2/2m
So this is an expression for kinetic energy of a particle of mass m and charge q moving in a circular path of radius r in magnetic field B.
The charge and the mass of a particle are important factors in this energy expression. For the two particles in the problems, we can compare those quantities qualitatively (without numbers) by a knowledge of what an alpha particle is:
proton: mass = m, charge = e
alpha: mass = 4m, charge = 2e
This is because an alpha particle is a helium nucleus, containing two protons (each with mass m and charge e) and two neutrons (each with, effectively, mass m, and no electric charge).
For the proton, we can then write the KE expression for the circular path as:
KEp = e2B2r2/2m
For the alpha particle, as I read the problem, we assume that the magnetic field B and path radius r are the same as for the proton. The kinetic energy for that particle then becomes:
KEa = (2e)2B2r2/2(4m) = 4e2B2r2/8m = e2B2r2/2m
This is exactly the same expression as for the proton. So the alpha particle turns out to have the same kinetic energy, 8 eV, in this case. This comes from the specific circumstance of having increased the mass by a factor of 4, but the charge only by a factor of 2.
I hope this helps! If you want to discuss any more about this, or look at the details, just let me know.