We are given the water density $\rho_w = 1.0 \times 10^3\,\text{kg/m}^3$, the container base area $A = 2.0 \times 10^{-2}\text{m}^2$, the submerged volume of the object $V_{sub} = 1.8 \times 10^{-3}\,\text{m}^3$, and the water level rise $\Delta h = 6 \text{ cm} = 0.06 \text{ m}$. We use the acceleration due to gravity $g = 10 \text{ m/s}^2$.

[Q1] The increase in pressure at the bottom of the container The increase in hydrostatic pressure at the bottom is due to the rise in the water level, $\Delta h$. The formula for the change in hydrostatic pressure is:

$$ \Delta p_{bottom} = \rho_w g \Delta h $$

Substituting the given values:

$$ \Delta p_{bottom} = (1.0 \times 10^3\,\frac{\text{kg}}{\text{m}^3})(10\,\frac{\text{m}}{\text{s}^2})(0.06\,\text{m}) = 600\,\text{Pa} $$

[Q2] The buoyant force acting on the object According to Archimedes' principle, the buoyant force, $F_B$, is equal to the weight of the fluid displaced by the object. The volume of displaced fluid is the submerged volume of the object, $V_{sub}$.

$$ F_B = \rho_w g V_{sub} $$

Substituting the given values:

$$ F_B = (1.0 \times 10^3\,\frac{\text{kg}}{\text{m}^3})(10\,\frac{\text{m}}{\text{s}^2})(1.8 \times 10^{-3}\,\text{m}^3) = 18\,\text{N} $$

[Q3] The increase in pressure the container exerts on the horizontal table surface The increase in the force that the container system exerts on the table is equal to the weight of the object, $W_{obj}$, added to the container.

$$ \Delta F_{table} = W_{obj} $$

Since the object is floating, its weight is balanced by the buoyant force, so $W_{obj} = F_B$. The increase in pressure on the table is this force divided by the container's base area, $A$.

$$ \Delta p_{table} = \frac{\Delta F_{table}}{A} = \frac{W_{obj}}{A} = \frac{F_B}{A} $$

Using the expression for $F_B$ from Q2:

$$ \Delta p_{table} = \frac{\rho_w g V_{sub}}{A} $$

Substituting the values:

$$ \Delta p_{table} = \frac{18\,\text{N}}{2.0 \times 10^{-2}\,\text{m}^2} = 900\,\text{Pa} $$