Millimole of $\mathrm{H}_2 \mathrm{SO}_4 \rightarrow 10 \times 2$

So Millimole of $\mathrm{NH}_3=20 \times 2=40$

Organic $\rightarrow$  $\mathrm{NH}_3$

Compound  40 Millimole

$\therefore$ Mole of $\mathrm{N}=\frac{40}{1000}$

wt. of $\mathrm{N}=\frac{40}{1000} \times 14$

$\%$ composition of $\mathrm{N}$ in organic compound

$=\frac{40 \times 14}{1000 \times 1} \times 100$

$=56 \%$