“n” and “m” are the primary and secondary valency of the
element ‘E’ in the complex $\left[E{\left(en\right)}_2\left(C_2{O}_4\right)\right]N{O}_2$  (where (en) is ethylene diamine) respectively, then $\frac{m}{n}$  is 
 

In the given complex we have two bidentate ligands (i.e en and  ), so coordination number of E is 6   (2 × 2 + 1 × 2 =6).
Let the oxidation state of E in complex be x, then [x + (-2) = 1] or x – 2 = 1 or x = +3, so its oxidation state is +3