Throughput is the amount of useful data sent per unit time. Useful data means payload.
Since the window size is $6$, the sender can send a maximum of $6$ frames till the first frame is acknowledged.
Time is taken by the first frame to get acknowledged= transmission time of original frame + propagation delay from src to destination + transmission time of ack frame + propagation delay from receiver to src.
Since ack transmission is to be neglected, we get total time$= 70+300+300 = 670 micro sec$.
The data sent during this time $= 6 frames= 6*1200bytes = 7200 bytes$.
Hence, throughput achieved is $7200 bytes/670 micro sec = 85.97Mbps$.