A hollow conducting cube A of walls of very small thickness at temperature T emits radiation like a black body. The cube A is kept in a room of temperature T0(

The rate of heat loss by cube of thickness Δ x , side  ℓ and made of density ρ  is given by −msdTdt=σA(T4−T04) ⇒−dTdt=σA(T4−T04)ms⇒−dTdt∝A⇒−dTdt∝ℓ2Rate of cooling is directly proportional to surface area.If the side of cube is doubled, surface area becomes four times. And thus, rate of cooling also becomes four times.