Answer: -3100 J

Volume of 1 mol of water at 373 K, 1 atm ($V_1$) = $18 cm^3$ = 0.018 L

Volume of 1 mol of water in vapour state at 373 K, 1 = $\frac{RT}{P} = \frac{0.0821 cm^3 atm K^{-1} mol ^{-1} \times 373 K}{\text{1 atm}} = 30.62 $ L

Now $w = -P(V_2 - V_1)$ = 1(30.62 - 0.018) = -30.6 L-atm = -30.6 $\times$ 101.3 = -3100 J (Hydrated salt)