Let the size of the transport layer packet i.e size of datagram payload be $x$.
Connection Establishment Overhead = 100 Bytes
Disconnection Overhead = 28 Bytes.
Total overhead = 100 + 28 = 128 Bytes
Now transport layer wants to keep an overhead to a minimum of 12.5%
So 12.5% of x = 128
$\frac{12.5}{100} * x = 128$
$x=$ $\frac{128*100}{12.5} = 1024 Bytes$
Option D is correct.