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Dear Student, 

The reaction involve in this : $N_2 +\hspace{0.17em}{O}_2\to 2NO$
& the no. of moles of $N_2, O_2,\hspace{0.17em}\& NO is 1,2 \& 3 respectively$. and the volume of solution given 100 litres. 
Concentration of $N_2, O_2,\hspace{0.17em}\& NO is 0.01, 0.02 \& 0.03 M respectively$. 
So, the equilibrium constant $$\begin{gathered} K = \frac{[NO{]}^2}{\left[N_2\right]\left[O_2\right]} \\ = \frac{(0.03{)}^2}{0.01\times 0.02} \\ = 4.5 \end{gathered}$$

Now, if the concentration of NO is changed to 0.04 , even though K value will be same. So, the concentration of oxygen can calculated. Let the new concentration of oxygen will be 'x'
                            ​$$\begin{gathered} 4.5=\frac{0.04\times 0.04}{0.01\times x} \\ x = \frac{0.04\times 0.04}{4.5\times 0.01} \\ = 0.0356 M \end{gathered}$$
Now the number of oxygen will be = 0.0356 x 100 = 3.56 moles of oxygen.
So, the extra moles of oxygen to be added = 3.56 - 2 = 1.56 mole or 100/64 mole.

So, the answer will be (iv) None of these.

Regards