A

Step 1: Determine the depth of water acting on the box. The lid of the box is at a depth of 18.6 m.
Step 2: Calculate the pressure at this depth using the formula:
$$ P = ho g h $$
where:
- $$ \rho $$ (density of water) = 1000 kg/m\(^3\)
- $$ g $$ (acceleration due to gravity) = 9.81 m/s\(^2\)
- $$ h $$ (depth) = 18.6 m
Thus,
$$ P = 1000 \times 9.81 \times 18.6 $$
Step 3: Calculate the upthrust (buoyant force). The box is 1 m x 1 m x 3 m high, thus the volume displaced by the box is 3 m\(^3\). The upthrust can be calculated using the formula:
$$ F_b = P \times A \times h $$
where A (area of the base) = 1 m x 1 m = 1 m\(^2\). Therefore,
$$ F_b = P \times 1 \times 3 $$
Step 4: Substitute for P:
$$ F_b = (1000 \times 9.81 \times 18.6) \times 1 \times 3 $$
Step 5: Therefore,
the upthrust acting on the box is approximately 548,358 N.
Therefore, upthrust can be determined to be significant, affirming that A is correct.