Percentage of oxygen=22.53

Precentage of metal=100-22.53 = 77.47

Formula of the metal oxide=MO

Let the atomic mass of the metal= x

Percentage by weight of the metal in the oxide MO=x *100/x+16=77.47

77.47x + 1239.52 = 100x

22.53x = 1239.52

x = 55.02 gm

Thus, the atomic mass of the metal is 55 gm approximately.

Percentage of oxygen in the second oxide = 30.38%

So, percentage of metal = 100 - 30.38 = 69.62%.

Now, follow the table - 

Element      Precentage      Atomic Mass         Atomic ratio           Simplest ratio       whole no.ratio

M                69.62                 55                 69.62/55=1.27         1.27/1.27=1               1x2 = 2

O               30.38                  16                 30.38/16 = 1.9         1.9/1.27=1.5            1.5x2 = 3

So, it becomes M₂O₃, the formula of the second oxide.