Volume in Cemistry Best Answer, Im a little lost

Determine the thickness (in cm) of a rectangular piece of aluminum foil that is 17.0 inches long and 3.0 inches wide, with a mass of 2.21 grams. Its density is 2.70 g/cm^3

d = m / V

V = m / d
V = 2.21 g / 2.70 g/cm^3
V = 0.818518519 cm^3

Let x = thickness of piece of aluminum

17 inches -----> 43.18 cm
3 inches -------> 7.62 cm

V = 43.18 cm * 7.62 cm * x cm = 0.818518519 cm^3
329.0316 x cm^3 = 0.818518519 cm^3
x = 0.818518519 / 329.0316
x = 0.002487659

Rectangular piece of aluminum is approx. 0.0025 cm thick

density = mass / volume
now b*w*h = v