The correct order of decreasing basic strength of the given amines is:

The correct order of decreasing basic strength of the given amines is:
N-ethylethanamine > ethanamine > N-methylaniline > benzenamine

To understand this order, let us analyze the basic strength of aliphatic and aromatic amines step-by-step:

1. Aliphatic Amines vs. Aromatic Amines:
Aliphatic amines (like N-ethylethanamine and ethanamine) are significantly stronger bases than aromatic amines (like N-methylaniline and benzenamine). In aliphatic amines, the alkyl groups are electron-donating groups due to the inductive effect (+I effect), which increases the electron density on the nitrogen atom, making its lone pair more readily available for protonation.
In contrast, in aromatic amines, the lone pair of electrons on the nitrogen atom is delocalized into the benzene ring through resonance. This resonance stabilization makes the lone pair less available for donation to an acid (proton).

2. Comparing Aliphatic Amines (N-ethylethanamine vs. Ethanamine):
* N-ethylethanamine ($(C_2{H}_5{)}_{2}NH$) is a secondary aliphatic amine. It has two ethyl groups.
* Ethanamine ($C_2{H}_{5}N{H}_2$) is a primary aliphatic amine with one ethyl group.
The +I effect of two ethyl groups in N-ethylethanamine increases the electron density on the nitrogen atom more than the single ethyl group in ethanamine. Additionally, in aqueous medium, secondary ethyl-substituted amines are more basic than primary amines because the combined effects of inductive influence and hydration energy outweigh steric hindrance. Thus:
$\text{N-ethylethanamine}>\text{ethanamine}$

3. Comparing Aromatic Amines (N-methylaniline vs. Benzenamine):
* N-methylaniline ($C_6{H}_{5}NHC{H}_3$) has a methyl group attached to the nitrogen atom.
* Benzenamine (Aniline) ($C_6{H}_{5}N{H}_2$) has only hydrogen atoms attached to the nitrogen in addition to the phenyl ring.
The methyl group in N-methylaniline is electron-donating (+I effect). This increases the electron density on the nitrogen atom relative to benzenamine, making N-methylaniline a stronger base than benzenamine. Thus:
$\text{N-methylaniline}>\text{benzenamine}$

Conclusion:
Combining the two comparisons, we get the overall order of decreasing basic strength:
$\text{N-ethylethanamine}>\text{ethanamine}>\text{N-methylaniline}>\text{benzenamine}$