Answer is 97.
Let's see MTU =200B
So we can send maximum 200B at once.
Here header size is 20B
So we can send only 200-20=180B at once but we cannot send data of size not divisible by 8.
We need to send 176 of data + 20 byte in one packet
So number of fragments=17076/176=97.022727
97 fragments.
But wait, there is some fraction part left (0.022727)
Let's break it down
If we send 97 fragments with each 176 bytes, in total we are sending =97*176=17072bytes
But total data is 17076bytes i. e we are left with 4byte data. Should we send this with an extra fragment?
Answer is NO.. because we are sending 196byte data in each fragments but the MTU is 200byte and as it is last byte we can send the remaining data along with this.
So the last fragment I.e fragment no. 97 will contain total 200 byte of data.
1st fragment 0---175(176B+20B)
2nd fragment 176---351(176B+20B
.....
96th fragment 16720---16895(176B+20B)
97th fragment 16896---17076(180B+20B)
If data was given atleast 1 byte extra then we might have to use another fragment for this.