As there are 480 registers, the bits require to represent registers is 9(log 480).
Now as we have 2 register operands we are left with 6 bits(24-18) for opcode.
The number of 2 address instructions possible = 2^6.
But we are only using 48 instructions from the 64, 2 address instructions.
2 address instructions not used = 64 - 48 = 16.
Total number of 1 address instruction possible = 16 * (2^9) => 8192
Let x be total number of 1 address instruction we are using.
So total number of 1 address instruction not used = 8192 - x
Now total number of 0 address instructions possible = (8192 - x) * 2^9
(8192 - x)*2^9 = 2048
x= 8188.
Total number of 1 address instructions possible = 8188