The meaning of the question isn't clear.
I assume the mass of the box is 3330kg, so its weight in Newtons is 3330g N, where g is the acceleration of gravity. If the density of water is about 1 g/cm3=0.001/(0.01)3=0.001/0.000001=1000kg/m3, then, since the weight of water displaced by the box must also be 3330g N, the volume V is given by density=mass/volume=3330/V=1000, so V=3.33m3.
The base area of the box=LB=4.7×1.9=8.93m2, so, if the box is to float, the depth D of the base below the surface of the box in the water is given by LBD=3.33m3, D=3.33/8.93=0.373m. The upward force of the water counterbalancing the weight of the box is 3330g Newtons (g=acceleration of gravity).
If D=0.38m then LBD=3.3934m3, which is equivalent to a mass of 3393.4kg. This is 63.4kg more than the actual mass of the box. If the mass of the box is increased by this amount, it will float with 0.38m below the surface. If a force of 63.4g Newtons is applied to anywhere except over the centre of gravity of the box, the box will tip over.
(The density of water is closer to 997.8kg/m3, but this would make V=3.34m3 instead of 3.33m3 and the floating depth would be 0.374m instead of 0.373m. This is still less than the given depth of 0.38.)