Here $(2357)_8$ is an octal number representation in which each digit can be represented as $\log_28=3$ bit binary number.
To convert an octal number to a hexadecimal number, first, write its binary equivalent in the group of $3$ bit.
$(2357)_8=(\ 010 \ 011 \ 101 \ 111)_2$
Now we can rearrange this binary number into the group of $4$ bits to get the hexadecimal number.
$(\ 010 \ 011 \ 101 \ 111)_2=( \ 0100 \ 1110 \ 1111)_H =(4EF)_H$
$\therefore (2357)_8=(4EF)_H$
Option $C$ is correct.
Gate 2004