Two bodies, A and B , which possess equal surface areas, are...

According to Stefan-Boltzmann law, the thermal radiation emitted by a body per unit time is given by $E = \sigma A T^4$

Given temperatures are $T_A = 10^{{}^{{}^{\circ}}}C = 283$ K and $T_B = 20^{{}^{{}^{\circ}}}C = 293$ K

The ratio of thermal radiation emitted is

$\dfrac{E_A}{E_B} = \left(\dfrac{T_A}{T_B}\right)^4 = \left(\dfrac{283}{293}\right)^4$

$\dfrac{E_B}{E_A} = \left(\dfrac{293}{283}\right)^4 = \left(1 + \dfrac{10}{283}\right)^4$

Calculating the value, we get \left(\dfrac{293}{283}\right)^4 \approx 1.149 \approx 1.15$<br><br>Therefore, the ratio is$1 : 1.15