A little game with rolling a fair dice You play the following game: you roll a fair dice then either you stop rolling and take the sum of the rolled numbers so far or you continue rolling the dice. Any time if a 1 is rolled you lose all your money, and has no option to continue the game. The strategy that you follow is that you wait until the cumulated price reaches a given level and then you stop. What should be this level in order to maximize the expected value of the prize?

My answer is 20...is it correct?
Let's say that $k$ points is the place where rolling again and stopping have the same expected value.  That means that $$k=\frac16(0)+\frac16(k+2)+\frac16(k+3)+\frac16(k+4)+\frac16(k+5)+\frac16(k+6)\\k=\frac56k+\frac{20}6\\\frac16k=\frac{20}6\\k=20$$

So you are correct.
Every time you don't lose you will win an average of 4. Total prize would be
$$ W_n = 4 \cdot n \rightarrow_n \infty $$
The probability of winning is 5/6 and after n turns, the probability of still winning something is:
$$ P_n = ( \frac{5}{6} )^n \rightarrow_n 0 $$

Total prize expectancy is 
$$ W_n \cdot P_n = 4 \cdot n \cdot( \frac{5}{6} )^n $$

Which has two maxima at n equal to 5 and 6.
$$ W_5 \cdot P_5 = W_6 \cdot P_6 = \frac{4 \cdot 5^6 }{ 6^5 } \approx 8.04 $$

So my strategy would be to play for 5 or 6 turns what will maximize my earnings. Every time I win, I will earn an average of 20 or 24.

Nice game to play with friends. Game fee 10$ ;)