Step-by-Step Solution:

1. Identify the Physical Principles The problem involves heat transfer between three parallel plates. Since the plates are separated, the primary mode of heat transfer is radiation. The system is in a steady state, which means the temperature of the middle plate is constant. For this to happen, the rate of heat energy absorbed by the middle plate must be equal to the rate of heat energy it emits. The plates are ideal black surfaces, so their emissivity is e = 1. The power radiated by a black surface of area A at temperature T is given by the Stefan-Boltzmann law: $P = \sigma A T^4$, where $\sigma$ is the Stefan-Boltzmann constant.

2. Set up the Energy Balance for the Middle Plate Let the plates be numbered 1, 2, and 3. Their temperatures are $T_1 = 2T$, $T_2$ (to be found), and $T_3 = 3T$, respectively. Let A be the area of each plate.

   • Heat Absorbed by the Middle Plate (Plate 2): Plate 2 is situated between Plate 1 and Plate 3. It absorbs radiation from both plates. Since the plates are large and parallel, Plate 2 absorbs all the radiation emitted by Plate 1 towards it and all the radiation emitted by Plate 3 towards it.

     • Power absorbed from Plate 1: $P_{1 \to 2} = \sigma A T_1^4$
     • Power absorbed from Plate 3: $P_{3 \to 2} = \sigma A T_3^4$
     • Total power absorbed by Plate 2 is the sum: $P_{absorbed} = P_{1 \to 2} + P_{3 \to 2} = \sigma A T_1^4 + \sigma A T_3^4$

   • Heat Emitted by the Middle Plate (Plate 2): The middle plate has two surfaces, one facing Plate 1 and the other facing Plate 3. It radiates energy from both surfaces.

     • Power emitted from the surface facing Plate 1: $P_{2, left} = \sigma A T_2^4$
     • Power emitted from the surface facing Plate 3: $P_{2, right} = \sigma A T_2^4$
     • Total power emitted by Plate 2 is the sum: $P_{emitted} = P_{2, left} + P_{2, right} = 2 \sigma A T_2^4$

3. Apply the Steady-State Condition In the steady state, the rate of energy absorption equals the rate of energy emission for Plate 2. $P_{absorbed} = P_{emitted}$ $\sigma A T_1^4 + \sigma A T_3^4 = 2 \sigma A T_2^4$

4. Solve for the Temperature of the Middle Plate ($T_2$) We can cancel $\sigma A$ from both sides of the equation: $T_1^4 + T_3^4 = 2 T_2^4$ $T_2^4 = {{T_1^4 + T_3^4} \over 2}$

5. Substitute the Given Temperatures We are given $T_1 = 2T$ and $T_3 = 3T$. $T_2^4 = {{(2T)^4 + (3T)^4} \over 2}$ $T_2^4 = {{16T^4 + 81T^4} \over 2}$ $T_2^4 = {{97T^4} \over 2}$

6. Calculate the Final Expression for $T_2$ Take the fourth root of both sides: $T_2 = {\left( {{{97T^4} \over 2}} \right)^{1/4}}$ $T_2 = {\left( {{{97} \over 2}} \right)^{1/4}} T$

7. Compare with Options The calculated temperature $T_2 = {\left( {{{97} \over 2}} \right)^{1/4}} T$ matches option (C).

Therefore, the temperature of the middle plate under steady state condition is ${\left( {{{97} \over 2}} \right)^{1/4}}T$.