If an approximation is not good enough, the definition of "standard deviation" is "The square root of the mean of the difference from the mean, squared". That is $\sqrt{\frac{\sum (x- \mu)^2}{n}}$ where $\mu$ is the mean and n is the number of terms. (Who ever gave you this problem probably expected you to know that!)
Here, the data set is {11, 12, 13, 14, 15}. The mean is $\frac{11+ 12+ 13+ 14+ 15}{5}= \frac{65}{5}= 13$. Notice that this is the "middle" number in the set. That works because this is an "arithmetic sequence".
Now, subtract that mean from each number and square:
11- 13= -2 and squaring, 4.
12- 13= -1 and squaring, 1.
13- 13= 0 and squaring, 0.
14- 13= 1 and squaring, 1.
15- 13= 2 and squaring, 4.
The mean of those numbers is $\frac{4+ 1+ 1+ 4}{5}= \frac{10}{5}= 2$.
The standard deviation is $\sqrt{2}$.
Now do the same with the altered data set, {9, 12, 13, 14, 17}, decreasing the smallest number, 11, by 2 to get 9 and increasing the largest number, 15, by 2 to get 17. The mean is $\frac{9+ 12+ 13+ 14+ 17}{5}= \frac{65}{5}= 13$ again. (Think about why that is true.)
Now subtract that mean from each number and square:
9- 13= -4 and squaring, 16.
12- 13= -1 and squaring, 1.
13- 13= 0 and squaring, 0.
14- 13= 1 and squaring, 1.
17- 13= 4 and squaring, 16.
The mean of those numbers is $\frac{16+ 1+ 0+ 1+ 16}{5}= \frac{34}{5}= 6.8$.
The standard deviation Is $\sqrt{6.8}= \sqrt{4(1.7)}= 2\sqrt{1.7}$.
The initial standard deviation was $\sqrt{2}$ which is approximately 1.414 while the new standard deviation is $\sqrt{6.8}$ which is approximately 2.608. The standard deviation has increased by approximately 2.608- 1.414= 1.914