@Magma I also knew the same before but i do get notifications :P
And here we have to subtract 0x10FEB2019 from each of the options and the result should be a multiple of 8.
Memory is by default taken as byte addressable. So one address corresponds to 1 byte.
64 bits instructions means each instruction is 8B long. So program counter will increment by 8B i.e. by the size of each instruction. Since 1 address corresponds to 1 byte so 8 addresses will correspond to 8B. So each instruction occupy 8 addresses. For going from instruction 1 to instruction 2 we have to skip 8 addresses.
I have not calculated the answer because i take to long to so these calculations :P
