Draw a diagram.

Consider the triangle made up by the VOR, the aircraft's 0020 position and the aircraft's 0035 position.

The angle at the VOR = 50º + 40º = 90º.

Since the two sides VOR-0020 and VOR-0035 are both 40nm it is an isosceles triangle and the other two angles are 45º each. At 0020 the bearing from the aircraft to the VOR is 310º - 180º = 130º so the aircraft's track is 130º - 45º = 085º. Since the four answers are 085º, 090º, 080º and 088º it must be 085º - 226 kt.

There is absolutely no need to work out the groundspeed.

You can work out the ground speed by considering the VOR-0020-0035 triangle again.

It is a right angle triangle and we can work out the 0020-0035 distance by trigonometry = 40/sin45º = 56.6nm.

This could also be done using Pythagoras.

This distance has been covered in 15 minutes so the groundspeed is 226 kts.

The problem is that we still have two possible answers and still have to work out the track angle.

Top tip on a question like this is to calculate the unique value (in this case track) and not the duplicated value (in this case groundspeed) because it will save time.