We recall the Ampere's circuital law for magnetic field.
Around a cylindrical cable (long one) the magnetic field B is along the tangential direction. This can be known from right hand thumb rule. B will be equal all along the circular path. So integral on LHS = B * 2 π d , d = distance of point from center of cylindrical wire.
On the RHS , i = algebraic sum of all currents enclosed (going out/coming in) with in the circular path.
If there was only one conductor wire: then B * 2 π d = μ₀ I
So we know B.
As there are two conductors, and currents are in opposite direction, RHS = 0. Hence B = 0 outside both conductor wires.
In between two conductor wires, circular path 2...
RHS = I as we do not include the current of the outer cable.
so B = μ₀ I / (2 π d)
Inside the inner cylindrical wire:
let radius of inner cable = R
let the radius of circular path chosen be = d
amount of current passing inside that circle: I * π d²/ π R²
so B * 2 π d = μ₀ I * d² / R²