Consider the two instances between which we're considering the change in velocity.
The first instance the speed is 6.0 m/s, and the direction is northerly.
The stopper then moves around one-ninth of a turn - i.e., 20 degrees. The speed is still 6.0 m/s. We therefore have two vector quantities. Bring them together, and use trig to find the change in velocity.
I'll admit this is a little unintuitive, which is why I wont continue the question to completion in case I misplace the angle when setting up my vector diagram. Usually any question like this will be such that the two instances are at right angles to each other, and so it's simple pythagoras and relatively easy to determine; I doubt you'd see this kind of question on any VCAA endorsed assessment.
Sorry I can't conclude the question (someone with spesh knowledge might be more helpful) but I hope this points you in the right direction.