$${F_3 = \frac{Mg}{4} - \frac{dAE}{4L} \\ = \frac{(290\, \mathrm{kg})(9,8\, \mathrm{m/s^2})}{4} \\ - \frac{(5,0 \times 10^{-4}\, \mathrm{m})(10^{-4}\mathrm{m^2})(1,3 \times 10^{10}\, \mathrm{N/m^2})}{(4)(1,00\, \mathrm{m})} \\ = 548\, \mathrm{N} \approx \underline{\underline{5,5 \times 10^2\, \mathrm{N}}}}$$