d
(d) ${(Q)_{Black\,body}} = A\sigma {T^4}t$
==> $\frac{Q}{t} \propto $$P = A\sigma {T^4}$
Breadth are halved so area becomes one fourth.
==> $\frac{{{P_1}}}{{{P_2}}} = \frac{{{A_1}}}{{{A_2}}} \times {\left( {\frac{{{T_1}}}{{{T_2}}}} \right)^4}$
==> $\frac{{{A_1}}}{{({A_1}/4)}} \times \left( {\frac{{273 + 327}}{{273 + 127}}} \right)$
==> ${P_2} = \frac{{81}}{{64}}E$