The ratio of emissive power of perfectly black body at 1327\ ^ ^ C...

According to Stefan-Boltzmann law, the emissive power of a perfectly black body is given by $E = \sigma T^4$, where $T$ is the absolute temperature.

Converting the given temperatures to Kelvin:
T_1 = 1327^{}^{\circ}\text{C} + 273 = 1600\text{ K}
T_2 = 527^{}^{\circ}\text{C} + 273 = 800\text{ K}

The ratio of their emissive powers is:
$\dfrac{E_1}{E_2} = \left(\dfrac{T_1}{T_2}\right)^4$

Substituting the values:
$\dfrac{E_1}{E_2} = \left(\dfrac{1600}{800}\right)^4 = (2)^4 = 16$

The ratio is $16 : 1$.

Answer: $16 : 1$