I'll outline some things but avoid explicit solution. The point is for you to try to work it out.
At each step, you will have some amount of points, $M$. If you stand for the next roll you will either win $k$ more points (giving you $M+k$) or you will lose $M$ (leaving you with $0$).
The expected value-maximizing strategy would say that any time a $5/6$ chance of $k$ is worth more than a $1/6$ chance of $M$ -- i.e. when the expected gain from playing would be $\frac56 k - \frac16 M$ is positive -- you benefit (on average) from continuing. A little mental arithmetic lets you see when to sit down for a variety of similar games of this type (as long as the game doesn't become much, much more profitable later, there's no need to sum series, you just stay standing until the net gain on the next roll is no longer positive).
HOWEVER, if you are trying to be the person with the most points of all at the end, maximizing expected value each turn is NOT the optimum strategy to be the eventual winner!
(For example, imagine at the end of the semester, you're coming second but the person coming first is a fair way ahead of you. If you sit down at the point that would maximize your expected return for that single round, you may simply make it certain you don't win.)
If you were to be playing many more times you would (approximately) seek to maximize expected value per turn, but as you get toward the last few rounds the winning strategy changes (if you're behind, you should take more risk, if you're ahead, less risk).
You can work out the exact strategy mathematically, but my guess is that the teacher is trying to get you to maximize your expected winnings (which works quite well in the early to mid stages of the iterated game).
You might like to look at articles on strategy for the dice game Pig (of which this is a variant). There's lots of information on strategy to be found, but its more fun to work it out in detail yourself.