$dT = \left( {3{T_0} - 2{T_0}} \right) = {T_0}$

Tmperature of surrounding, ${T_{surr}} = {T_0}$

Initial temperature of sphere, ${T_{initial}} = 3{T_0}$

Specific heat of the material of the sphere varies as,

$c = \alpha {T^3}\,per\,unit\,mass\,\left( {\alpha  = a\,constant} \right)$

Applying formula,

$\frac{{dT}}{{dt}} = \frac{{\sigma A}}{{McJ}}\left( {{T^4} - T_{surr}^4} \right)$

$ \Rightarrow \frac{{{T_0}}}{{dt}} = \frac{{\sigma 4\pi {R^2}}}{{M\alpha {{\left( {3{T_0}} \right)}^3}\,J}}\left[ {{{\left( {3{T_0}} \right)}^4} - {{\left( {{T_0}} \right)}^4}} \right]$

$ \Rightarrow dt = \frac{{M\alpha 27T_0^4J}}{{\alpha 4\pi {R^2} \times 80T_0^4}}$

Solving we get,

Time taken for the sphere to cool down temperature $2{T_{{0^,}}}$

$t = \frac{{M\alpha }}{{16\pi {R^2}\sigma }}In\left( {\frac{{16}}{3}} \right)$