On doubling the initial concentration, the half life of a reaction becomes half. The order of the reaction is

2

The correct answer is Option (3) → 2

For a reaction, if initial concentration is doubled and the half-life becomes half, this matches the behavior of a second-order reaction.

For a second-order reaction:

$t_{1/2} = \frac{1}{k[A]_0}$

So, if $[A]_0$​ is doubled, $t_{1/2}$​ becomes half.