The ammonia ( NH _ 3 ) released on quantitative reaction of 0.6g...

${NH}_{2}{CONH}_{2} + 2NaOH \to 2{NH}_{3} + {Na}_{2}{CO}_{3}$

By, stoichiometry, $1mol$ of urea gives $2mol$ ${NH}_{3}$

Moles of urea =$= \dfrac{0 . 6gm}{60gm / mol} = 0 . 01mol$

So, $0 . 01$ mole urea produce $0 . 02$ mole ${NH}_{3}$

To neutralize ammonia $HCl$ is required

${NH}_{3} + HCl \to {NH}_{4}Cl$

By equation, it is clear that $1mol$ of ammonia is neutralized by 1mol of $HCl$.

So, $0 . 02mol$ of ammonia neutralized by $0 . 02mol$ of $HCl$

$∴ MoleofHCl = 0.02mole$

$EquivalentofHCl = normality \times volume = 0 . 2N \times 100mL = 0 . 02equivalent$

As $n$-factor of $HCl$ is $1$, so $numberofequivalents = numberofmoles = 0 . 02moles$

Hence, option $3$ is correct.