Interesting problem. By abuse of notation, let the lengths of the ships P, Q, and R also by P, Q, and R, respectively. I chose variables to denote lengths rather than areas or volumes to avoid square or cubic roots. Then the problem says that Q + R = 2P, i.e., P is the arithmetic mean of Q and R. Since we are also told that Q is the bigger of the three ships, we have R < P < Q.

Since the number of crew is proportional to the deck area, which itself is proportional to the square of length, the crews of P, Q, and R are , and , respectively, for some unknown coefficient . The problem tells us that , or .

Since the volume grows faster with linear size than the number of crew, it is better to transfer cargo from Q to as many intermediate-sized P's as possible and use the smallest R's only for the remainder. So let us find Q / P. From the two equations above, Q / P = 3 / 2 in my calculations (needs double-checking). Similarly, Q / R = 3 (same remark). Therefore, the cargo of Q can be transferred to 1 P and 1 R.