Henry’s constant (in kbar) for four gases $\alpha ,\beta ,\gamma and \delta$ and  in water at 298 K is given below

(density of water = 103kgm–3 at 298 K)
This table implies that :

Given, Henry's constant
We need to find This table implies that, from the given option.

$$\begin{gathered} 1. P_{\gamma }=K_H{X}_Y \\ \begin{array}{l}{\mathrm{P}}_{\gamma }=2\times {10}^{-15}\times \frac{55.5}{55.5+\frac{1000}{18}}=2\times {10}^{-5} \mathrm{K}\text{ bar }\\ =2\times {10}^{-2}\text{ bar }\end{array} \\ \text{ (2) }{\mathrm{P}}_{\delta }={\mathrm{K}}_{\mathrm{H}}{\mathrm{X}}_{\delta } \\ {\mathrm{P}}_{\mathrm{\delta }}=0.5\times \frac{55.5}{55.5+\frac{1000}{18}}=.249 \mathrm{Kbar}=249\mathrm{bar} \\ \left(3\right) \mathrm{As} \mathrm{temperature} \mathrm{rises}, \mathrm{gases} \mathrm{become} \mathrm{less} \mathrm{soluble}. \\ \text{ (4) }{\mathrm{K}}_{\mathrm{H}}\downarrow \text{ solubility }\uparrow \text{ and lowest }{\mathrm{K}}_{\mathrm{H}}\text{ is for }\mathrm{\gamma } \end{gathered}$$

Therefore, the correct option is A The pressure of a 55.5 molal solution of $\delta$ is 250 bar