Other’s have explained very well but i’m just making more deeper insight.
On the right side of equality we have even number, while on left we have to add 3 numbers. It seems an easy task, but the numbers given to be used are all odd. We clearly know that adding 3 odd numbers won’t give us an even result. So we have to make these three boxes contain even number. The best and easiest we could do is to make all contain 10. ( i.e 10+10+10=30)
How could we have an even number in 3 boxes? Just read the question, there is no restriction of using other operations within these boxes. So one way to do this is,
(1+9)+(3+7)+(5+5) = 30
while another way is already suggested by Sneha Krishnam. There are a lot ways, just play with several other operations.