Let’s assume the output of $1^{st}$ multiplexer is Z.
now according to the output equation of $4\times1$ multiplexer:
$Z=\bar U \bar VI_0+\bar UVI_1+U\bar VI_2+UVI_3$
$\implies Z=0+\bar UV+U\bar V+0)$
$I_0$ ,$I_3$ is connected to ground & $I_2$,$I_4$ is connected to $V_{cc}$
$\left[{\because V_{cc}=1,Ground =0}\right]$
$\implies Z= \bar UV+U\bar V$
$\implies Z=(U\oplus V)$
Now $F=\bar W\bar XI_0+\bar WXI_1+W\bar XI_2+WXI_3$
$\implies F=\bar W\bar X(U\oplus V)+\bar WX(U\oplus V)$
$\implies F=(U\oplus V)(\bar W\bar X+\bar WX)$
$\implies F=(U\oplus V)(\bar W(X+\bar X))$
$\implies F=(U\oplus V)\bar W$
Option C is correct.