The problem is merely your understanding of floating point numbers and the methods used to store them.
MATLAB uses only 52 binary bits of precision to store a floating point number.
What would it require to store that number?
log2(str2sym('2.00000000000000000000000000095861') - 2)
So 100 binary bits would just barely suffice to approximate that number, but not that closely. And as I said, MATLAB stores a floating point number as a DOUBLE, thus only 52 binary bits. There is no 16 byte floating point class, and even that would probably be barely sufficient. And of course, if you had such a long precision form available, we would constantly see questions just like yours...
"Why won't MATLAB exactly represent this number?
The answer is to understand how a number is stored and how to work with that. This sometimes requires a good grounding in numerical methods.