The ammonia (NH₃) released on quantitative reaction of 0.6 g urea (NH₂CONH₂) with sodium hydroxide (NaOH) can be neutralized by

NH₂CONH₂ + 2NaOH $\to$2NH₃ + Na₂CO₃
From this, 1 mole urea gives 2 mole ammonia.
Therefore, moles of urea = 0.6/ 60 = 0.01 mole; moles of ammonia = 2x0.01 = 0.02 mole
Similarly, by balancing the other reaction, 0.02 mole NH₃ requires 0.02 mole HCl.
We know, for 0.2 M HCl,
$$\begin{gathered} Molarity = \frac{Moles }{Volume} \\ 0.2 = \frac{Moles of HCl}{{\displaystyle \frac{100}{1000}}} \\ Moles of HCl = 0.02 mol HCl \end{gathered}$$
Hence, correct option is: (A) 100mL of 0.2 N $\mathrm{HCl}$