Here, my friend. do it as follows :-

 

1.      Reaction is   :    2N₂0₅   ->  4NO₂ + O₂

2.      Now, at STP, 1 mole of gas occupies 22.4 litres.

3.      So, 28 litres of oxygen are occupied by 1.25 moles of O₂.

4.      Now, you have provided the molarity as well as the volume 

         of  N₂0₅ present at the starting through which you can calculate the

         initial amount of moles present.

5.      Also, 2 moles of N2O5 gives 1 mole of oxygen.

        So, 1.25 moles of oxygen is given by 2.5 moles of N2O5.

6.     So, now you get the amount of N2O5 left

7.     As far as rate is concerned, it is  = -- ( Amount of N2O5 used ) / time