Think of a piece of playdough; if you roll it to be twice as long, it must become much thinner, making it harder for 'current' to flow through the now restricted, narrow path.

R is proportional to L squared (when stretching): $R \propto l^2$.

The resistance of a wire is given by $R = \rho \frac{l}{A}$. Since the volume $V = A \times l$ is constant, stretching the wire to $n$ times its length ($l' = nl$) reduces the cross-sectional area to $A' = \frac{A}{n}$. Substituting these into the formula, $R' = \rho \frac{nl}{A/n} = n^2 \rho \frac{l}{A} = n^2 R$.

Always check if the problem specifies 'volume remains constant' before applying the $n^2$ rule; if a new wire is simply cut, only $l$ changes and $R$ changes linearly.