First, consider the equivalent thermal resistance $R_{\text{eq}}$ of the series arrangement:

$ R_{\text{eq}} = R_1 + R_2 + R_3 $

Calculating each resistance:

$ R_1 = \frac{1}{2KA}, \quad R_2 = \frac{1}{KA}, \quad R_3 = \frac{1}{2KA} $

Summing these gives:

$ R_{\text{eq}} = \frac{1}{2KA} + \frac{1}{KA} + \frac{1}{2KA} = \frac{2}{KA} $

In a series arrangement, the rate of heat flow is constant. Thus:

$ \frac{3T - T_1}{R_1} = \frac{3T - T}{R_{\text{eq}}} $

Substituting the resistances:

$ \frac{(3T - T_1) \cdot 2KA}{I} = \frac{2T \cdot KA}{2I} $

Solving the equation yields:

$ 6T - 2T_1 = T \implies 2T_1 = 5T \implies T_1 = \frac{5T}{2} \quad \ldots (1) $

Next, examine the heat flow rate in the third section:

$ \frac{T_2 - T}{R_3} = \frac{3T - T}{R_{\text{eq}}} $

Substitute:

$ \frac{(T_2 - T) \cdot 2KA}{I} = \frac{2T \cdot KA}{2I} $

Solving gives:

$ 2T_2 - 2T = T \implies 2T_2 = 2T + T \implies T_2 = \frac{3T}{2} \quad \ldots (2) $

By substituting equations (1) and (2) into the ratio:

$ \frac{T_1}{T_2} = \frac{5T/2}{3T/2} = \frac{5}{3} $

Thus, the ratio $\frac{T_1}{T_2}$ is $\frac{5}{3}$.