An organic compound ‘A’ on treatment with NH₃, gives ‘B’  which on heating gives ‘C’, ‘C’ when treated with Br₂, in the  presence of KOH produces ethylamine. Compound ‘A’ is:

Since, C when heated with Br₂, in presence of KOH produces ethylamine, hence it must be propanamide and hence the organic compound (A) will be propanoic acid. The reactions follows. 
$\underset{\left(\mathrm{A}\right)}{{\mathrm{CH}}_3-{\mathrm{CH}}_2-\mathrm{COOH}\stackrel{{\mathrm{NH}}_3}{⟶}}\underset{\left(B\right)}{{\mathrm{CH}}_3-{\mathrm{CH}}_2-{\mathrm{COONH}}_4\stackrel{\mathrm{\Delta }}{⟶}}$         $\underset{\left(\mathrm{C}\right)}{{\mathrm{CH}}_3 -{\mathrm{CH}}_2- {\mathrm{CONH}}_2} \underset{\mathrm{Hoffmann} \left(\mathrm{C}\right) \mathrm{bromamide} \mathrm{reaction} }{\overset{\mathrm{KOH}+\mathrm{Br}2}{\to }}$$\underset{(\mathrm{Ethylamine} ) }{{\mathrm{CH}}_3-{\mathrm{CH}}_2-{\mathrm{NH}}_2}$