The correct answer is C. 6W.
Time Division Multiplexing (TDM) is a technique for combining multiple data streams into a single stream by assigning each stream a separate time slot. The minimum bandwidth required for transmission of a TDM signal is the sum of the bandwidths of the individual data streams.
In this case, the four messages are band limited to W, W, 2W and 3W respectively. Therefore, the minimum bandwidth required for transmission of the TDM signal is $W + W + 2W + 3W = 6W$.
Option A is incorrect because it is the bandwidth of only one of the four messages. Option B is incorrect because it is the sum of the bandwidths of two of the four messages. Option D is incorrect because it is the sum of the bandwidths of three of the four messages.