The answer must be A.
(Using $\wedge$ to denote bitwise AND)
For 1$^{st}$packet,
$(128.75.43.16) \wedge (255.255.255.0) = (128.75.43.0)$ since $\{16 \wedge 0 = 0\}$, as well as
$(128.75.43.16) \wedge (255.255.255.128) = (128.75.43.0)$ since $\{16 \wedge 128 = 0\}$.
Now, since both these subnet masks are producing the same Network ID, hence The one with greater number of ones will be selected, and the packet will be forwarded there. Hence packet $1$ will be forwarded to Eth$1$.
For $2^{nd}$ packet,
$(192.12.17.10)$ when ANDed with each of the subnet masks does not match with any of the network ID, since:
$(192.12.17.10) \wedge (255.255.255.0) = (192.12.17.0)$ {Does not match with any of the network addresses}
$(192.12.17.10) \wedge (255.255.255.128) = (192.12.17.0)$ {Does not match with any of the network addresses}
$(192.12.17.10) \wedge (255.255.255.255) = (192.12.17.10)$ {Does not match with any of the network addresses}
Hence, default interface must be selected for packet $2$, i.e., Interface Eth$2.$