We do not write out the subnet mask bits in binary, we have a way to write out subnet masks when entering them on network equipment and when writing them out by hand.

If we steal five host bits from the third octet we have to add the binary values together

128	64	32	16	8	4	2	1
1	1	1	1	1	0	0	0

So we have 128+64+32+16+8 = 248

Remember that we are using a class B example here and so are working with the third octet.  We are not allowed to alter the first two octets, they are fixed.