Let the speed of the plane be p Kmph.
So the speed of plane from A to B will be 'p+50' and the speed from B to A will be 'p-50'.
We notice that the plane goes from B to A stays there for 1 hr and again come back to B with total time duration 12 hrs.
So we have $$\frac{3000}{p-50} + 1 + \frac{3000}{p+50} = 12$$.
We can clearly see that speed of the plane is 550 which satisfies the above equation.
So from going to A from B the plane takes $$\frac{3000}{550-50} = 6$$ hrs.
So time at B when plane reaches at A is 2 pm .
Hence the time difference between A and B is 1 hr.

Alternatively,
Let speed of flight be s,
Since A is to the east of B, A is ahead of be in time
Let A be ahead of B in time by a hours
Departure from A = 4PM, Arrival at B = 8PM
Travel time = 8 - 4 +a = 4 + a
Since City B is behind city A by 'a' hours, the actual travel time is 'a' hours more than the difference of local times.
Similarly when one travels from B to A, since B is ahead of A by 'a' hrs, actual travel time is 'a' hours less than total
i.e. B->A Travel time = (3PM - 8AM) - a = 7 -a
Total distance travelled = Speed $$\times$$ Time taken ....(1)
From A to B, the wind is favourable / in same direction as flight
Hence from (1), we have 
A->B >>> $$3000 = (s+50)(4+a) => 3000(7-a) = (s+50)(4+a)(7-a)$$ ...(2)
B->A >>> $$3000 = (s-50)(7-a) => 3000(4+a) = (s-50)(7-a)(4+a)$$ ...(3)
(2) - (3) => $$3000(3-2a) = 100(7-a)(4+a) => a^2-63a+62=0 => a=1/62$$
Hence the time difference between A and B is 1 hr.