Time complexity of $\sim c \cdot n^3$ with a computer which is 10 times faster
You've correctly reasoned that the additional speed doesn't change the $n^3$ part of the tilde expression. The only component left for it to effect then is $c$. 

Let $c_1$ be the constant of the first computer and $c_2$ the constant of the second. If the second computer is 10x faster, then $c_2 = c_1/10.$ Now, all you need to do is solve the following equation:

$$c_1 \cdot n^3 = \frac{c_1 \cdot n'^3}{10},$$

which gives the same answer you came up with!