I solved it this way:
For the first instruction format, the opcode is of 6 bits. So, this gives $2^6$ possible combinations. But, 3 of these patterns represent other instruction formats, therefore $2^6 - 3$.
Now for each of these 3 special patterns, the opcode is specified in some other 6 bits of the instruction. So, for each of these 3 instructions, we have $2^6$ opcodes possible, i.e. $3 * 2^6$
This gives us the total answer:
$2^6 - 3 + 3 * 2^6$
= 61+192
= 253