B

Step 1: Understanding Electric Field Inside Conductors
In electrostatics, the electric field inside a conductor in electrostatic equilibrium is zero. This applies to the inner regions of both the smaller and larger spherical shells when you are within their conducting material.

Step 2: Analyzing the Charge Distribution
1. The inner shell has a total charge of +3Q. By Gauss's law, a charge enclosed within a Gaussian surface will contribute to the electric field outside that surface. Since the charge is on the outer surface of the inner spherical shell, the electric field inside the shell is zero.
2. The outer spherical shell has a total charge of +5Q. The charge on the inner surface of the larger shell will be equal and opposite to that of the inner shell enclosed; therefore, it will have a charge of -3Q on its inner surface, resulting in a net positive charge on its outer surface (+8Q).

Step 3: Matching the Options
Column-I:
(i) Electric field strength is Zero - This corresponds to a region inside the conducting material of any conducting shell. This matches with [B] for the inner surface of the larger spherical shell (r is between the surfaces X and A).
(ii) Electric field strength is non-zero - This matches with [D] for the region between the outer surface of the smaller shell and the inner surface of the larger shell (a < r < b).
(iii) Magnitude of charge on this surface is 3Q - This corresponds to the outer surface of the smaller shell, which retains the charge due to the inner charge distribution. This matches with [C].
(iv) Charge on this surface is +8Q - This corresponds to the outer surface of the larger shell which has +8Q total charge. This matches with [A].

Conclusion:
Based on the analysis, the matching pairs are:
(i) - [B]
(ii) - [D]
(iii) - [C]
(iv) - [A]
Therefore, the correct answer is B.