Neither of you is answering the question as asked. It is not the change of long which is 20º, but the earth convergence.
Convergency = change of long x sine mean lat
20 = ch long x .866
ch long = 20/.866 = 23.09º
(This is itself a slight approximation, because the mean lat will actually be higher than 60N, but not by much).
This means that there is 11.547005º ch long (or chart convergence) between the start point and the mid-point of the straight-line Polar Stero track. Using SWH's method, this gives a co-latitude distance at the mid-point of the straight-line track of 1763.569 nm or, if you prefer, a latitude of N6036.43. (There is a slight inaccuracy caused because the scale expansion of a Polar Stereographic is not linear, but a function of the secant squared of half the co-latitude, which will give a very slightly different scale expansion between the N Pole and N6000 and the N Pole and N6036, but I think that we can afford to ignore that).
We now need to work out the latitude of the Great Circle track at mid-point, and this is best solved by spherical trigonometry.
Draw a triangle with A as the North Pole, B as N 6000 W 01132.82, and C as the mid-point of the Great Circle track. Label the opposite sides to angles A,B, and C as sides a, b and c.
Angle A = 11.5470005º
Angle C = 90º
side c = 30º (co-latitude of B)
sin a/ sinA = sin c/sin C
This gives side a as 5.7441º of Great Circle arc (344.65 nm, if you're interested).
We can now use Napier's rules:
sin b = tan a x co-tangent A
So side b (which is the co-latitude of the mid-point of the Great Circle) is 29.4953º, or a co-latitude distance of 1769.72 nm.
This gives the latitude of point C as N6030.28
We now know the lat of the midpoint of the straight line - N6036.43 - and the latitude of the mid-point of the Great Circle - N6030.28. So the cross track-displacement is 6.15 nm.
So none of the given answers is correct.
Are you sure you got the question right, WhiteKnight? It is not,as far as I know, a JAA question, yet you give 4 options as answers.
I suspect that it should have been the rhumb-line v straight-line cross track distance, that it's the change of longitude, not the convergency which is 20 degrees, and that this isn't a real JAA question but an FTO worksheet example.