Since the hard disk has a $16$-bit data bus, it can transfer two bytes at one time. Thus its transfer rate is $50/2 = 25 \: K$ half-words ($16$-bits each) per second. This corresponds to an overhead of $20$ instructions or $25 \: K \times 20 =25 \times 2^{10} \times 20 = 512000$ instruction per second.