Step 1: Find the size of the original IP packet.
The UDP payload is $7488$ bytes, and the UDP header adds another $8$ bytes.
So, the total data handed over to IP is:
\[ 7488 + 8 = 7496 \text{ bytes} \]
Since the IPv4 header has no options, its size is $20$ bytes.
Therefore, the total size of the original IP datagram becomes:
\[ 7496 + 20 = 7516 \text{ bytes} \]
Step 2: Determine how much data each fragment can carry.
The maximum transmission unit (MTU) of Ethernet is $1500$ bytes, which includes both IP header and data.
After accounting for the IP header, the usable data per fragment is:
\[ 1500 - 20 = 1480 \text{ bytes} \]
This value is suitable for fragmentation since it is a multiple of 8, as required by IPv4.
Step 3: Calculate the number of fragments.
The total IP data that must be fragmented is $7496$ bytes.
The number of complete fragments is:
\[ \left\lfloor \frac{7496}{1480} \right\rfloor = 5 \]
The remaining data for the final fragment is:
\[ 7496 - (5 \times 1480) = 96 \text{ bytes} \]
Thus, the total number of fragments formed is:
\[ 5 + 1 = 6 \]
Step 4: Find the size of the last fragment.
The last fragment carries only $96$ bytes of data.
Including the IP header, its total size is:
\[ 96 + 20 = 116 \text{ bytes} \]
Step 5: Match with the given choices.
Options with fewer than 6 fragments are incorrect.
Options with an incorrect final fragment size do not match the calculation.
The only consistent choice is the one stating 6 fragments with the last fragment size of 116 bytes.
Step 6: Final conclusion.
The correct result is:
\[ \boxed{\text{6 fragments, last fragment size } = 116 \text{ bytes}} \]