Problem 900

If the product of three numbers $x$, 36 and $y$ is 2592, then $x \cdot 36 \cdot y = 2592$ and so $xy = \dfrac{2592}{36} = 72$.
If $x$ and $y$ are positive integers with $xy = 72$, then we have the following possibilities:

$x$
$y$
$x + y$

72
1
73

36
2
38

24
3
27

18
4
22

12
6
18

9
8
17

We have assumed that $x>y$ since we have not assigned an order to $x$, 36 and $y$.
In the given problem, we want to put four pairs of numbers in the outer circles so that the 9 numbers are different and the sum of the 9 numbers is as large as possible.
Putting this another way, we want to choose 4 of the 6 pairs in the table above (knowing that we cannot choose the pair 36 and 2 since 36 is already in the middle circle) to make the sum as large as possible.
Since we know the sums of the pairs, we choose the pairs with the four largest sums.
This means that the sum of the 9 numbers will be $(72+1)+(24+3)+(18+4)+(12+6) + 36$ which equals $73 + 27 + 22 + 18 + 36$ or $176$.