==> $P \propto {T^4}$
Also energy received per sec (p)$ \propto \frac{1}{{{d^2}}}$
(inverse square law)

==> $P \propto \frac{{{T^4}}}{{{d^2}}}$

==> $\frac{{{P_1}}}{{{P_2}}} = {\left( {\frac{{{T_1}}}{{{T_2}}}} \right)^4} \times {\left( {\frac{{{d_2}}}{{{d_1}}}} \right)^2}$

==> $\frac{P}{{{P_2}}} = {\left( {\frac{T}{{2T}}} \right)^2} \times {\left( {\frac{{2d}}{d}} \right)^2} = \frac{1}{4}$

==>${P_2} = 4P.$