Equivalent current due to rotation of charged ring,
I = q f = (2πrσ)n
Magnetic field in the axis of ring, at a distance ‘y’ from its center

If we consider a circle of radius 'r' with the center at the center of given circle, then different points on this circle are at different distances from the given wires. Thus magnetic field at different points will be different. We can find the field if the position of point is specified, but we cannot determine the field at any general point from given data.

Note that, here Ampere's law is not useful.
If we consider the above circle to be an Amperian loop, then net current encircled by the loop will be = I + I - 2 I = 0
So from ampere circuital law

As we are given that the wire is same on all sides, so we can say that resistance of each side of triangle will be equal. 
So resistance of the section ABC will be = 2R, And resistance of section AC will be = R 
Thus current in section ABC

Magnetic field at P due to wires (1) and (2) is:

Now if a current of  is flowing in the third wire, then magnitude of magnetic field at point P due to it will be

If the current is from right to left, then field due to this wire will be into the screen, and thus it will cancel out the field due to other two wires.

Let us first consider the forces and torques acting on the sphere.

For translational equilibrium along the incline, there should be friction acting on the sphere, up the incline plane.

Torque of friction, normal reaction and mg cos θ will be zero wrt point P.

So for rotational equilibrium of the sphere about point P,

torque due to mg sin θ should be balanced by torque due to magnetic force

(mg sin θ) R = M B sin θ

m g R = i ( π R² ) B

For equilibrium about O 
m g ( R cos θ ) - T₂ ( 2R cos θ ) - B i π R² cos θ = 0
m g - T₂ ( 2 ) - B i π R = 0
mg - 2 T₂ - η mg = 0 ( Given : π B i R = η mg )

For same speed, r ∝ m
So we can say that proton will follow the bigger circle.

However, we cannot determine the direction of revolution of particles, because depending in the initial direction of velocity, any particle can revolve clockwise or anti-clockwise.

For example, if both particles are projected upwards, then proton will revolve anti-clockwise and electron will revolve clockwise.
If electron is projected downwards, then it will also revolve anti-clockwise.

As field due to an "out of screen” current will be anti-clockwise, so the integral for it will be positive. So we have to take an "out of screen” current to be positive and an "into the screen” current to be negative.

The required values will simply be proportional to the current encircled by the loops with proper signs.

loop A = μ₀ (4i - 2i) = 2 μ₀ i 
loop B = μ₀ (3i - 0) = 3 μ₀ i 
loop C = μ₀ (i - 2i) = - μ₀ i 
loop D = μ₀ ( - i) = - μ₀ i 

So, B > A > C = D