According to Henry's law, the partial pressure of a gas in vapour phase ($p$) is proportional to the mole fraction of the gas ($x$) in the solution, and is expressed as $p = K_H x$. This implies that at a constant pressure, the higher the value of Henry's law constant ($K_H$), the lower is the solubility of the gas in the liquid .

Given $K_H$ values: Ar = 40.3 kbar $CO_2$ = 1.67 kbar $CH_4$ = 0.413 kbar HCHO = $1.83 \times 10^{-5}$ kbar

Since the decreasing order of $K_H$ is Ar > $CO_2$ > $CH_4$ > HCHO, the increasing order of solubility will be the exact reverse: Ar < $CO_2$ < $CH_4$ < HCHO.