They have two ice bars. The dimensions of the first bar are 3dm * 3dm * 3dm, the dimensions of the second

They have two ice bars. The dimensions of the first bar are 3dm * 3dm * 3dm, the dimensions of the second are 4dm * 3dm * 2dm. The mass of which of the bars is less and by how much and at what time?

a1 = 3 dm = 0.3 m.
a2 = 4 dm = 0.4 m.
l1 = 3 dm = 0.3 m.
l2 = 3 dm = 0.3 m.
h1 = 3 dm = 0.3 m.
h2 = 2 dm = 0.2 m.
m1 / m2 -?
The mass of a substance m is determined by the formula: m = ρ * V, where ρ is the density of the substance, V is the volume of the substance.
m1 = ρ * V1, m2 = ρ * 2.
The volume of the bar V is found as the volume of a rectangular parallelepiped by the formula: V = a * l * h, where a is the width, l is the length, h is the height of the bar.
V1 = a1 * l1 * h1, V2 = a2 * l2 * h2.
m1 = ρ * a1 * l1 * h1.
m2 = ρ * a2 * l2 * h2.
m1 / m2 = ρ * a1 * l1 * h1 / ρ * a2 * l2 * h2 = a1 * l1 * h1 / a2 * l2 * h2.
m1 / m2 = 0.3 m * 0.3 m * 0.3 m / 0.4 m * 0.3 m * 0.2 m = 1.125.
Answer: the mass of the first bar is 1.125 times greater than the mass of the second bar m1 / m2 = 1.125.