A solid copper sphere of radius 0.010 m is painted black (treat it as a black body)…

Step 1. The sphere. Its surface area is $4\pi r^2 = 12.566\ \text{cm}^2$, its mass $37.28\ \text{g}$, and its heat capacity $mc = 14.35\ \text{J/K}$.

Step 2. The heater (a). The sphere radiates $\sigma A T^4$ and takes in $\sigma A T_0^4$ from the walls. The heater makes up the difference:

Step 3. The first rate (b). $\dfrac{dT}{dt} = -\dfrac{P}{mc} = -\dfrac{8.66}{14.35}\ \text{K/s} = -0.603\ \text{K/s}$.

Step 4. The time (c). Separate the variables in $mc\,\dfrac{dT}{dt} = -\sigma A(T^4 - T_0^4)$:

Here $\dfrac{mc}{\sigma A} = 2.0144 \times 10^{11}\ \text{K}^4\,\text{s}$, and with the given formula ($a = 300\ \text{K}$) the integral is $1.2420 \times 10^{-9}\ \text{K}^{-3}$. So $t = 250\ \text{s}$.

Step 5. A larger sphere (d). The heat capacity grows as $r^3$ and the area as $r^2$, so $mc/\sigma A$ grows as $r$. The same cooling takes twice as long: $500\ \text{s}$.