The correct option is C 24
357678=(3574)169×(3572)
This can be re-written as,
(574)169×(572)
since we are concerned only with the last two digits.
Now, (572×572)169×(572)
=(49×49)169×49
=(01)169×49
=01×49=49
Finally, 49 when divided by 25 will leave a remainder of 24.