What question says is
100 elements are already present in the stack. Say in the bottom up manner from 1,2,3, . . . 100.
Now to remove 1 we need to do 100 pop operation and then save remaining 99 elements to temporary location (as not specified to save it to another stack) and then again push those 99 elements to stack again.
So now the stack has elements 2,3,4,. . .,100. Now to remove 2 we again follow the same procedure so we will end up with 99 pop operations followed by 98 push operations.
Continuing this to 10 times we will get no of $\textbf{PUSH = }$ 99 + 98 + . . . + 90 = 945