A 0.85 g sample is a mixture of urea, CO(NH2)2 (molar mass 60 g mol^-1), and glucose. In a Kjeldahl determination all the nitrogen is converted to…

The acid taken is $50.0\ \mathrm{mL}\times0.25\ \mathrm{M}\times2=25.0$ milliequivalents of $\mathrm{H^+}$ because sulfuric acid is dibasic. The back-titration consumes $18.0\times0.50=9.0$ milliequivalents of base, so the excess acid is $9.0$ meq. The ammonia absorbed neutralised $25.0-9.0=16.0$ meq of acid, so $n(\mathrm{NH_3})=16.0\ \mathrm{mmol}$. This equals the nitrogen content: $m(\mathrm{N})=16.0\times10^{-3}\times14=0.224\ \mathrm{g}$. Glucose contains no nitrogen, so all of it comes from urea, which is $\dfrac{28}{60}$ nitrogen by mass. Hence $m(\text{urea})=0.224\times\dfrac{60}{28}=0.480\ \mathrm{g}$. The percentage is $\dfrac{0.480}{0.85}\times100=56.47\%$. The key points are that the dibasic acid counts two equivalents per mole, that only the ammonia-consuming portion of the acid is relevant after back titration, and that the nitrogen-to-urea conversion factor $60/28$ is used because urea carries two nitrogen atoms per molecule.