We know that,

22400 lit of nitrogen = 1 mole of nitrogen = 28 gm of nitrogen

So, 5.6 lit = 28x5.6/22400

= 0.007 gm.

Weight of oxide = 11 gm

So, the oxygen component of the oxide = 11 - 0.007

= 10.993 gm

Similarly, weight  of second oxide = 15 gm.

Here, weight of oxygen = 10.993 gm, 

so weight of nitrogen = 15 - 10.993

= 4.007 gm

So, the amount of nitrogen reacting with a fixed amount of oxygen (10.993 gm) is -

1. in the first case 0.007 gm

2. in the second case 4.007 gm 

So they are in ratio 0.007:4.007 

= 1:572

It is in a ratio of whole numbers, thus supports the law of multiple proportion. 

However, the question may have some mistakes, because a ration of 1:572 is not expected between nitrogen and oxygen. So kindly check the question.