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To determine the least acidic compound among the given choices, we compare the stability of the conjugate bases formed after the loss of a proton ($$H^+$$). The greater the stability of the conjugate base, the stronger the corresponding acid.
Among the given compounds, $$m$$-ethoxyphenol is a phenol. Phenols are acidic with a typical $$pK_a$$ of about $$10$$ because the resulting phenoxide ion is stabilized by resonance over the benzene ring.
Benzenesulfonic acid is a sulfonic acid and is a very strong acid with $$pK_a < 0$$. Its conjugate base, the sulfonate ion, is highly stabilized by resonance, with the negative charge delocalized over three oxygen atoms.
Pentanoic acid is a carboxylic acid with a typical $$pK_a$$ of approximately $$4.8$$. The corresponding carboxylate ion is resonance-stabilized by delocalization of the negative charge over two equivalent oxygen atoms, making it considerably more acidic than phenols.
Ethyl propiolate ($$EtO_2C-C \equiv CH$$) contains a terminal alkyne hydrogen. Although terminal alkynes are acidic due to the $$sp$$-hybridized carbon atom, their acidity is much lower, with a typical $$pK_a$$ around $$25$$. The electron-withdrawing ester group increases its acidity slightly, but it is still far less acidic than phenols, carboxylic acids, or sulfonic acids.
Therefore, the order of acidity is
$$\text{Benzenesulfonic acid} > \text{Pentanoic acid} > m\text{-Ethoxyphenol} > \text{Ethyl propiolate}.$$
Hence, the least acidic compound is
$$\boxed{\text{(D) Ethyl propiolate}}.$$
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