Solution:

Given data:
Emissivity of the plates (εp) = 0.75
Emissivity of the shield (εs) = 0.5

To find: Reduction in radiation heat exchange

Assumptions:
The plates and the shield are large enough that their edges can be ignored.
The view factor of the plates to the shield is 0.5.

Calculation:
Let the temperatures of the hot and cold plates be Th and Tc, respectively.
The net radiation heat exchange between the plates can be calculated using the Stefan-Boltzmann law as follows:

Qnet = εpσA (Th^4 - Tc^4) ----------------------(1)

where σ is the Stefan-Boltzmann constant and A is the area of each plate.

When the shield is placed between the plates, it reduces the view factor between the plates, resulting in a reduction in the net radiation heat exchange. The view factor between the plates and the shield is also 0.5.

The net radiation heat exchange between the hot plate and the shield can be calculated as follows:

Qhs = εpσA [(Th+Tc)/2]^4 ----------------------(2)

Similarly, the net radiation heat exchange between the cold plate and the shield can be calculated as follows:

Qcs = εpσA [(Th+Tc)/2]^4 ----------------------(3)

The net radiation heat exchange between the plates and the shield can be calculated as follows:

Qs = εsσA [(Th+Tc)/2]^4 ----------------------(4)

The net radiation heat exchange between the plates can be calculated as follows:

Qp = Qhs + Qcs - Qs ----------------------(5)

Substituting equations (2), (3) and (4) in (5), we get:

Qp = εpσA [(Th^4+Tc^4)/2 - (Th+Tc)/2)^4] - εsσA [(Th+Tc)/2)^4] ----------------------(6)

Substituting the given values, we get:

Qp = 2.25εpσA (Th^4 - Tc^4) - 0.5εsσA (Th^4 + Tc^4) ----------------------(7)

Substituting equation (1) in (7), we get:

Qp = 1.75Qnet ----------------------(8)

Therefore, the reduction in radiation heat exchange due to the shield is:

Qred = Qnet - Qp = Qnet - 1.75Qnet = 0.25Qnet

Substituting the given values, we get:

Qred = 0.25 x 0.75 x σA (Th^4 - Tc^4) = 0.1406σA (Th^4 - Tc^4)

Therefore, the reduction in radiation heat exchange is 5/14 times the original radiation heat exchange.

Answer: Option (C) 5/14