what Aditya is saying is correct but he made a slight mistake, if you see expanding opcode then the equation will be like [2^4-X]*2^2=8 from here if you solve you will get X=14.
Now, how you are getting the above equation –
for type I structure is 6|6|4 and for type R structure is 4|6|6
now if you know the concept of expanding opcode we always find the number of instruction which have highest bit opcode by going from lowest bit opcode to highest bit opcode
but here the question is tricky here they have given the no of instruction which has the highest bit opcode number.
so how we solve this is we will go from lowest to highest, we have 4-bit opcode in the R type so total instruction will be 2^4 and now let suppose X number of instruction are occupied, so the number of instruction which we can convert to I type is the remaining number of instruction i.e
2^4-X and now if you see the difference between the size of address bit in R and I that is 2bit (I – 6+4 = 10 and R→ 6+6=12, so we want to go from 12 to 10)
so we have to multiply that 2 bit in decoded form in the equation so [2^4-X]*2^2
and this is equal to I type of instruction which is 8
(PS:- I am not good at explaining stuff, and this concept is easy but the question is a little bit tricky)
Kindly refer this video to understand the concept correctly https://www.youtube.com/watch?v=6pWuV2mAt5E