physics question?

ρ = density of water = 1,000 kg/m³
V = volume (given) = 1.0 m³
m = mass of water = to be determined

ρ = m / V
ρ·V = m
m = ρ·V
m = 1,000 kg

Pressure is a force per unit area. In this case, the applied force is the weight of the water
since apparently the weight of the "very light plastic tank" can be assumed to be negligible.

A = area of base (given) = 1.0 m²
g = gravitational acceleration = 9.81 m/s²
W = weight of water = to be determined
P = pressure on ground = to be determined

W = m·g
W = 9,810 N

P = W / A
P = 9,810 N/m² = 9,810 Pa = 9.81 kPa

If we ignore the tanks mass, it's around 1000kg of water
Formula is: P=f/A A is 1 square meter(si) and f =mg=1000kg multiply to 9.8m/s^2( I have no idea what you call it, its g~9.8 so p=9800 pascal got it? Sorry for my poor English

If "light plastic" means an ideal zero-weight tank that maintains its structure and holds the water, (and not a light plastic which disintegrates) it is 1000 kg per square metre.

You have omitted the height of the water surface above the 1 m² base. If the tank tapers down toward the top, this will be greater than 1 m. The formula for water pressure is
P = ρ*g*h (Volume not required)