From what i understand from the question,
There are 1K microinstructions , a control word or a microinstruction size is 32 bits , and out of 1000 there are 450 unique or distinct micro instructions.
(A microinstruction is also a control word)
Hence size of micro programmed memory will be ---> 1024 * 32 = 32768 bits.
Now , we are using nano memory for which micromemory has indexing into nano memory. Micromemory will be used with 1K instructions and nanomemory contains 450 distinct instructions. micromemory will work whether whether a nano memory is used or not. But when used it will point to 450 unique instructions of the memory . This time bit width of control word or microinstruction will be log(450)= approx take as 9. To represent 450 nano memory words, we need 9 bits in micromemory .
so new micro memory size = 1024 * 9 = 9126 bits
and so formed nano memory size = 450 * 32 = 14400 bits
Hence total memory now will be = 9126 + 14400 = 23616 bits which was initially without nano memory = 32768 bits.
Total bits saved will be = 32768 - 23616 = 9152 bits
Total bits saved from micro memory will be = 32768 - 9126 = 23642 bits which is approx. 23kbits.
I hope this helps you , as no way we can decrease the nano memory size from 14400 bits = 14K bits