The order of basic strength of ${\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2\left(1\right),{\mathrm{C}}_2{\mathrm{H}}_5{\mathrm{NH}}_2\left(2\right),{\left({\mathrm{C}}_2{\mathrm{H}}_5\right)}_2\mathrm{NH}$ (3), and ${NH}_4$ (4) is;

 The basicity of aniline is lower than that of ammonia because the lone pair of electrons on $\mathrm{N}$ in aniline is involved in resonance with the benzene ring. Because of the presence of electron-releasing ethyl groups ( $+$ l effect), ethyl amine and diethyl amine are more basic than ammonia, and the lone pair of electrons on nitrogen can be easily transferred.

Thus, the correct order of basic strength is $-{\mathrm{C}}_6{\mathrm{H}}_5{\mathrm{NH}}_2<{\mathrm{NH}}_4<{\mathrm{C}}_2{\mathrm{H}}_5{\mathrm{NH}}_2<{\left({\mathrm{C}}_2{\mathrm{H}}_5\right)}_2\mathrm{NH}$.

Hence, the correct answer is option A) $1<4<2<3$.