Black body heat transfer

From the Stefan-Boltzmann Law of black-body radiation, we have the following result

$P = \sigma A (T^4 - T_c^4 ).$

where $P$ denotes the net radiated power, $A$ denotes the surface area of the blackbody , $T$ and $T_c$ denotes the temperatures of the black-body and surroundings respectively.

Now, $P$ can also be written as

$P = - \dfrac{\,dQ}{\,dt}$

where $\,dQ$ is the amount of heat radiated in time $\,dt$ (negative sign accompanies because of heat emission).

Also, we know that for preferentially solids, the amount of heat radiated $\,dQ$ can be expressed as

$\,dQ = mS \,dT.$

where $m$ is the mass of the body, $S$ is the specific heat capacity and $\,dT$ denotes the infinitesimal temperature change.

Putting together all our equations in one single equation, we get

$- \dfrac{mS \,dT}{dt} = \sigma A (T^4 - T_c^4 )$

Putting all the necessary information given and solving by integration, we get

$\begin{aligned} & - \dfrac{({4}/{3}) \pi R^3 \rho S \,dT}{\,dt} = \sigma (4 \pi R^2) (T^4 - 0) \\ \\ & \implies - \dfrac{R \rho S}{3} \cdot \dfrac{\,dT}{T^4} = \sigma \,dt \\ & \implies \displaystyle - \dfrac{R \rho S}{3} \int_{400}^{200} T^{-4} \,dT = \sigma \int_{0}^{t_f} \,dt \\ & \implies \dfrac{R \rho S}{9} \left( \dfrac{1}{200^3} - \dfrac{1}{400^3} \right) = \sigma t_f \\ & \implies t_f = \dfrac{R \rho S}{9 \sigma} \cdot \dfrac{7}{6.4 \times 10^7} \\ & \implies t_f = R \rho S \left( \dfrac{7}{9 \times 6 \times 10^{-8} \times 6.4 \times 10^7} \right) \\ & \implies t_f = \dfrac{175}{864} R \rho S \end{aligned}$

Thus $a+b = \boxed{1039}$ .