Two things must both be right — the conversion specifier and the ampersand — and only option 4 has both.

scanf("\n %d", &m);

Why %d. The specifier tells scanf how to interpret the incoming characters and how many bytes to write. %d means signed decimal integer, matching the declaration int m. Option 3's %c would store a single raw character — typing 5 would leave the code 53, not the value 5 — and option 2's %s would copy characters into memory as a string, overrunning a four-byte integer at once. Option 2 is also missing its closing quotation mark, so it will not even compile.

Why the ampersand. This is the real point of the question. C passes everything by value, so scanf receives copies of its arguments and could never change a caller's variable through one. It is given the address instead:

\(\texttt{\&m}\ \longrightarrow\ \text{a pointer to m}\)

and writes through that pointer. Omitting it, as option 1 does, passes the current value of m — whatever rubbish an uninitialised variable holds — and scanf treats that number as an address and writes there. The result is undefined behaviour: usually a segmentation fault, occasionally silent corruption of unrelated memory, which is worse.

The one exception to the rule is worth remembering because it is the source of much confusion: with %s and a character array, no ampersand is used — an array name already decays to the address of its first element. Everything else, including every numeric type, needs the &.

The leading \n in the format string is not required. Whitespace in a scanf format means "skip any amount of whitespace here", so "\n %d" and " %d" and "%d" behave identically for numeric input, which already skips leading blanks.

Hence, the correct form is scanf("\n %d", &m);.