Finding the mass of NaNO3 in original mixture from preciptate
The dried precipitate is $\ce{Ag2S}$. 
The number of moles of $\ce{Ag2S}$ is:
$$n=\frac{m_{\ce{Ag2S}}}{M_{\ce{Ag2S}}}=\frac{0.44}{32+108\times 2}=\frac{0.44}{248}=0.00177\,\mathrm{mol}$$
The number of moles of ion $\ce{Ag(I)}$ is$$n'=2n=0.00355\,\mathrm{mol}$$

On the other hand, the percentage by mass of $\ce{AgNO3}$ in the original mixture is: $$\frac{m_{\ce{AgNO3}}}{m_{\ce{(AgNO3 + NaNO3)}}}\times 100=\frac{n'\times M_{\ce{AgNO3}}}{m_{\ce{(AgNO3 +NaNO3)}}}\times 100=\frac{0.00355 \times (108+14+16 \times 3)}{1.61}\times 100=37.48\%$$

So, the percentage by mass of $\ce{NaNO3}$ is: $100-37.48=62.52\%$