We can see that $142.72.43.16$ bitwise AND $255.255.255.128$ gives $142.72.43.0 (128$ has only the MSB as $1)$
$ 142.72.43.16$ bitwise AND $255.255.255.0$ also gives $142.72.43.0.$
When multiple destinations match, the subnet with more number of $1s$ is chosen. So, here the first one - $\text{Eth1}$ wins.
$192.32.21.3$ bitwise AND $255.255.255.254$ gives $192.32.21.2$ - not matching any destination. So, default $\text{Eth4}$ wins.
So, correct answer: C.