Each Ethernet frame contains $1376\;\text{bytes}$ (largest multiple of $8 \leq 1380$) of user data and $20\;\text{bytes}$ of IP header. So for $5516\;\text{bytes}$ of user data and $8$ bytes of UDP header, we need $\left\lceil\dfrac{5516+8}{1376} \right\rceil = 5 $ IP fragments.
Offset of the last fragment $ = 4 \times 1376/8 = 688.$
So, $a+b = 5+688 = 693.$
Note: The Fragment Offset field $(13\;\text{bits})$ is used to indicate the starting position of the data in the fragment in relation to the start of the data in the original packet. This information is used to reassemble the data from all the fragments (whether they arrive in order or not). In the first fragment, the offset is $0$ as the data in this packet starts in the same place as the data in the original packet (at the beginning). In subsequent fragments, the value is the offset of the data the fragment contains from the beginning of the data in the first fragment $\text{(offset 0)}$, in $\textbf{8 byte 'blocks' (aka octawords).}$
