According to Henry's law, the partial pressure of a gas in the vapour phase ($p$) is directly proportional to its mole fraction ($x$) in the solution, 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 the solubility of the gas in the liquid .

Given the $K_H$ values: $K_H(\text{A}) = 145 \text{ kbar}$ $K_H(\text{B}) = 2 \times 10^{-5} \text{ kbar}$ $K_H(\text{C}) = 35 \text{ kbar}$

The decreasing order of $K_H$ values is A > C > B. Since solubility is inversely proportional to $K_H$, the decreasing order of solubility will be exactly the reverse: B > C > A.