15.03 cm³ CH₄(NTP) is dissolved in 500 cm³ water.

Thus, we have Volume v dissolved in 1 cm3 water = $\frac{15.30}{500}\times 1=0.03006$cm³

Amount of gas dissolved = $\frac{\mathrm{PV}}{\mathrm{RT}}=\frac{1\times 0.03006}{82.06\times 273}$

                                    =1.34 X 10^-6mol

Mass of gas disolved, m=1.34 X  10^-6 X 16 = 2.144 X  10^-5gm

Now, according to Henry's law, we have m=Kp (K- Henry constant. K=m/p.

Substituting the values of m and p, we get K=2.144 X 10⁵gm atm^-1

Now for 1 X  10^-3mole of gas in 300cm^-3 of water we have Mass of gas dissolved in 1 cm³ water

$=\frac{1\times {10}^{-3}\times 16}{300}\times 1=5.33\times {10}^{-5}\mathrm{gm}$

From Henry's law, pressure required to dissolve the above amount = $\frac{m}{k}=\frac{5.33\times {10}^{-5}}{2.144\times {10}^{-5}}=2.486\mathrm{atm}.$