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

The rate of heat loss from a body is given by the Stefan-Boltzmann law:

$\dfrac{dQ}{dt} = e \sigma A (T^4 - T_0^4)$

Since both spheres have the same emissivity $e$, are at identical temperatures $T$, and are in identical surrounding temperatures $T_0$, the rate of heat loss depends only on their surface areas $A$.

The surface area of a sphere is $A = 4 \pi R^2$, which implies $\dfrac{dQ}{dt} \propto R^2$.

Given that the copper sphere has twice the radius of the aluminium sphere, $R_{Cu} = 2R_{Al}$.

The ratio of the rate of heat loss from the aluminium sphere to that of the copper sphere is:

$\dfrac{(dQ/dt)_{Al}}{(dQ/dt)_{Cu}} = \dfrac{R_{Al}^2}{R_{Cu}^2} = \dfrac{R_{Al}^2}{(2R_{Al})^2} = \dfrac{1}{4}$

The specific heat capacity and density are extraneous information, as they are only required if the rate of cooling $\left(\dfrac{dT}{dt}\right)$ is asked.