Here, depth of water column,$h = 3\,\text{km} = 3 \times 10^{3}\,\text{m}$density of water,$\rho = 10^{3}\,\text{kg}\,\text{m}^{-3}$ If P be the pressure exerted by this water column at this depth, then$P = h \rho g = 3 \times 10^{3} \times 10^{3} \times 9.8$$= 29.4 \times 10^{6}\,\text{Pa}$$= 30 \times 10^{6}\,\text{Pa} = 3 \times 10^{7}\,\text{Pa}$As the structure is put on the sea, sea water will exert upward thrust of $3 \times 10^{7}\,\text{Pa}.$Maximum stress which the vertical off-shore structure can withstand $= 10^{9}\,\text{Pa}.$ (Given)Then $3 \times 10^{7}\,\text{Pa} < 10^{9}\,\text{Pa}$Thus we conclude that the structure is suitable as the stress applied by it is much lesser than the maximum stress it can withstand.