Amount of NaOH consumed= Amount of HCI unreacted

$n=VM=\left(15\times {10}^{-3}\mathrm{L}\right)\left(0.1\mathrm{molL}=15\times {10}^{-1}\mathrm{mol}\right.$

Amount of ${\mathrm{NH}}_3$ evolved =Amount of HCl reacted

= Amount of HCl taken to start with -Amount of HCl unreacted 

$\begin{array}{l}=\left(20\times {10}^{-3}\mathrm{L}\right)\left(0.1{\mathrm{molL}}^{-1}\right)-15\times {10}^{-4}\mathrm{mol}\\ =5\times {10}^{-4}\mathrm{mol}\end{array}$

Mass of ${\mathrm{NH}}_3$ evolved $=nM=\left(5\times {10}^{-4}\mathrm{mol}\right)\left(17\mathrm{g}{\mathrm{mol}}^{-1}\right)=85\times {10}^{-4}\mathrm{g}$

Mass per cent of nitrogen in the compound is

$\left(\frac{M_{\mathrm{N}}}{M_{{\mathrm{NH}}_3}}\right)\left(\frac{m_{{\mathrm{NH}}_3}}{m_{\text{compound }}}\right)\times 100=\left(\frac{14\mathrm{g}{\mathrm{mol}}^{-1}}{17\mathrm{g}{\mathrm{mol}}^{-1}}\right)\left(\frac{85\times {10}^{-4}\mathrm{g}}{29.5\times {10}^{-3}\mathrm{g}}\right)\left(100\right)=23.7$