Explanation

(d) According to Kirchhoff’s law the ratio of emissive power to absorptive power is same for all surfaces at the same temperature and is equal to the emissive power of a perfectly black body at that temperature
Hence, $\frac{e_{1}}{a_{1}} = \ \frac{e_{2}}{a_{2}} = \ldots\left\lbrack \frac{E}{A} \right\rbrack\ _{perfectly\ black\ body\ }$
Now, since ${(E\lambda)}_{black}$ is constant at a given temperature, this implies that good absorber is a good emitter (or radiator).