According to Henry’s Law ; $P_1=K_HX_1 $ and $P_2=K_HX_2$
$P_1=$1bar: $X_1=\large\frac{6.56\times 10^{-3}g}{30}$-----(i)
$P_2=?$
$X_2=\large\frac{5.00\times 10^{-2}g}{30}$-----(ii)
From eq(i) & eq(ii)
$\large\frac{P_1}{P_2}=\frac{X_1}{X_2}$
$\therefore P_2=\large\frac{5\times 10^{-2}}{6.56\times 10^{-3}}$
$P_2 = 7.62$ bar