Could drag in a turn be twice the drag of that encountered in straight and level flight? I Just wanted to see if my understanding of drag is correct, as I am unsure.
I understand that during straight and level flight, the drag can be calculated as:

With the first term being the parasitic drag component and the second term being the induced drag component.
However, in a coordinated turn with the same angle of attack at the same airspeed, I have been told that the total drag can be expressed as:

And because the angle of attack is the same, the lift to drag ratio is the same. In this scenario the load factor is 2.
However, since in straight and level flight L = W. So does it follow that the total drag in this turn is double the total drag in the straight and level flight?
i.e.

I don't know why but I feel that this is wrong.
If this is correct, however, the question says that the parasitic drag of the turn is equal to the parasitic drag of the straight and level flight; so would it correct to follow that the induced drag is equal to the following equation?
Could drag in a turn be twice the drag in straight and level flight?

Yes, but this will be a rater steep turn.

…  in a coordinated turn with the same angle of attack at the same airspeed …

No, this is impossible. Either angle of attack or speed have to be higher so the higher lift required for turning can be generated.

… angle of attack is the same, the lift to drag ratio is the same. In this scenario the load factor is 2

And in order to create twice the lift of straight flight, the speed has to be higher by a factor of √2. In this case, L/D is indeed the same, but speed is higher so twice the dynamic pressure is available for creating the required lift. Since both lift and drag are proportional to dynamic pressure, creating twice the lift at the same L/D means that also drag will double.
In order to turn with a load factor of 2g, the bank angle has to be 60°. That is rather steep.