Address field = 9 bits.
For two address instructions, 18 bits will be for address. Remaining (32-18) = 14 bits will be for opcode.
Using 14 bit opcode, 214 opcodes are possible. But there are only 400 two address instructions.
So, remaining (214-400) opcodes can be used for one address instructions.
Possible one address instructions = (214-400) * 29 [using opcode expand technique]