The sum of the lengths of the two shorter sides must be greater than the length of the longest side. With a perimeter of 20, the longest any side can be is 9.
Use the standard convention of having c as the longest side of the triangle, with a and b for the other two sides.
Make a list of all the sets of integers (order doesn't matter) with c the largest, c less than or equal to 9, and a+b+c = 20.
c a,b
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9 9,2
9 8,3
9 7,4
9 6,5
8 8,4
8 7,5
8 6,6
7 7,6
The condition for a right triangle is
The condition for an obtuse triangle is
To answer the question, test each of those sets of integers to find how many of them satisfy the condition for forming an obtuse triangle.
Note common sense tells us that, in an obtuse triangle, one of the sides is "much longer" than the other two sides. So it is likely, given the list we have, that the only obtuse triangles with integer side lengths and a perimeter of 20 will have one side with length 9.
You will find that is indeed the case.
I leave it to you to find the answer to the question.