A solid aluminium sphere and a solid copper sphere, the latter...

The rate of fall of temperature of a body is given by the Stefan-Boltzmann law:

$\dfrac{dT}{dt} = \dfrac{e \sigma A (T^4 - T_s^4)}{mc}$

For a solid sphere, the mass is m = \rho V = \rho \left(\dfrac{4}{3} \piR^3\right) and the surface area is $A = 4 \pi R^2$. Substituting these into the equation gives:

\dfrac{dT}{dt} = \dfrac{e \sigma (4 \pi R^2) (T^4 - T_s^4)}{\rho \left(\dfrac{4}{3} \pi R^3\right) c} = \dfrac{3 e \sigma (T^4 - T_s^4)}{\rho c R}

Since the temperatures, surrounding temperatures, and emissivities are identical for both spheres, the rate of fall of temperature $r$ is inversely proportional to the product of density, specific heat capacity, and radius:

$r \propto$\dfrac{1}{\rho c R}

The ratio of the rate of fall of temperature of the aluminium sphere to that of the copper sphere is:

\dfrac{r_{Al}}{r_{Cu}} = \dfrac{\rho_{Cu} c_{Cu} R_{Cu}}{\rho_{Al} c_{Al} R_{Al}}

Given values:
$R_{Cu} = 2 R_{Al}$
$\rho_{Cu} = 3.4 \rho_{Al}$
c_{Al} = 900 \text{ J kg}^{-1} ^\circ\textC^{-1}
c_{Cu} = 390 \text{ J kg}^{-1} ^\circ\textC^{-1}

Substituting these values into the ratio:

\dfrac{r_{Al}}{r_{Cu}} = \dfrac{3.4 \rho_{Al} \times 390 \times 2 R_{Al}}{\rho_{Al} \times 900 \times R_{Al}}

\dfrac{r_{Al}}{r_{Cu}} = \dfrac{3.4 \times 390 \times 2}{900}

\dfrac{r_{Al}}{r_{Cu}} = \dfrac{2652}{900} \approx$2.946$

Rounding to one decimal place, the ratio is approximately $2.9 : 1$.