The coefficient of linear expansion of steel is $\alpha=12 \times 10^{-6}/^{\circ} K$, so in coolling by $\Delta T=330^{\circ} C\ (=330^{\circ} K)$ the radius shrinks from $0.1\ \mbox{m}$ to $0.1-0.1\times \alpha \times \Delta T$ or by a fraction $\alpha \times \Delta T$, so the area reduces by about twice this as a fraction or $2 \times \alpha \times \Delta T\approx 2 \times 12 \times 10^{-6} \times 330 = 0.00792$, or $0.792 \%$.