Reaction of propanamide with Br$$_2$$ / KOH (aq) produces:

The reaction of an amide with bromine in the presence of aqueous potassium hydroxide is known as the Hofmann bromamide degradation (Hofmann rearrangement). In this reaction, an amide is converted into a primary amine containing one carbon atom fewer than the original amide.

The general reaction is

$$RCONH_2+Br_2+4KOH\rightarrow RNH_2+K_2CO_3+2KBr+2H_2O.$$

For propanamide ((CH_3CH_2CONH_2)), the carbonyl carbon is removed during the rearrangement, producing a primary amine with a two-carbon chain.

Thus,

$$CH_3CH_2CONH_2\xrightarrow{Br_2/KOH}CH_3CH_2NH_2.$$

Hence, the product formed is ethylamine.

Therefore, the correct answer is

$$\boxed{\text{Ethylamine }(CH_3CH_2NH_2)}.$$