Since the microprocessor has a 16-bit external data-bus, this means that 16 bits are transferred in 1 bus cycle.
now input clock= 8MHz
This means 8x(10^6) cycles in 1 second.
therefore, 1 cycle takes 1/(8x(10^6)) seconds
since 1 bus cycle= 4 clock cycles= 4/(8x(10^6)) seconds
now 1 bus cycle -------> 16 bits
4/(8x(10^6)) seconds ---------> (16x8x(10^6))/4 bits = (16x8x(10^6))/(4x8) BYTES
therefore answer= 4x(10^6) Bytes/sec