In this question , following courses of action will occur :
A) First prior to transmit the data , the sender needs to acquire the channel and we know for that ,
Time required = 2 * Propogation delay = 2 * 5 = 10 μs
B) Then data is placed into channel for that we need time = (Data + Header) / (Bandwidth)
= 256 / 10 *106
= 25.6 μs
C) Then the data goes through channel for that propogation delay = 5 μs
D) Then another 2 P.T. time is required by receiver to acquire the channel..That is what they are referring to as 'slots' ..So time taken for that = 10 μs
E) Now for placing acknowledgement , we need time = 32 / 10 * 106
= 3.2 μs
F) Now again propogation delay of ack is involved = 5 μs
Hence effective data rate = Data size ( excluding header and acknowledgement) / Total time involved as found above
= 224 / ( 10 + 25.6 + 5 + 10 + 3.2 + 5 )
= 3.8 Mbps