Answer
$$\left( 10^{5}+1030\times 10\times \dfrac {0.6}{2}\right) \times 0.6^{2}\approx 37112.4N$$
Work Step by Step
$$F=P\times A=\left( P_{atm}+\rho g\dfrac {L}{2}\right) L^{2}=\left( 10^{5}+1030\times 10\times \dfrac {0.6}{2}\right) \times 0.6^{2}=$$
$$\left( 10^{5}+1030\times 10\times \dfrac {0.6}{2}\right) \times 0.6^{2}\approx 37112.4N$$
