Answer
10 cm.
Work Step by Step
Hooke’s law says that the stretch is proportional to the applied force, $F \sim \Delta x$. The new force, 50 N, is $\frac{5}{3}$ of the old, 30 N. The spring will now stretch $\frac{5}{3}$ times as far, or 10 cm.
This is discussed on page 231.
