For maximum efficiency, the sender should be transmitting for the entire RTT (till ACK comes back).
RTT = Transmission Time for Packet + Propagation Time for Packet + Transmission Time for ACK (taking as 0) + Propagation Time for ACK
Transmission Time for Packet = Packet Size $\div$ bandwidth $=(64 \times 8) \div (1.536 \times 10^6) s = 333.33 \mu s.$
Propagation Time (same for Packet and ACK) $=3000 \; km \times 6 \mu s / km = 18000 \mu s$
So, RTT $ = 333.33 + 2 \times 18000 = 36333.33 \; \mu s$
So, No. of packets needed for $100\%$ efficiency $ = \lceil \frac{36333.33}{333.33} \rceil= 109.$
No. of bits needed to represent $109 = \log_2 109 = 7.$