Answer
$t \propto \frac{1}{\sqrt{\Delta l}}$ if $\Delta l \lt \lt l$
$t$ is constant if $\Delta l \gt \gt l$
Work Step by Step
We can find the tension in the rubber band:
$\tau = k~\Delta l$
We can find the wave speed:
$v = \sqrt{\frac{\tau}{\mu}}$
$v = \sqrt{\frac{k~\Delta l}{m/(l+\Delta l)}}$
$v = \sqrt{\frac{k~\Delta l~(l+\Delta l)}{m}}$
We can find an expression for $v$ if $\Delta l \lt \lt l$:
$v = \sqrt{\frac{k~\Delta l~(l+\Delta l)}{m}} \approx \sqrt{\frac{k~\Delta l~(l)}{m}}$
We can find an expression for the time if $\Delta l \lt \lt l$:
$t = \frac{l+\Delta l}{v}$
$t \approx \frac{l}{\sqrt{\frac{k~\Delta l~(l)}{m}}}$
$t \approx \sqrt{\frac{m~l}{k~\Delta l}}$
Therefore, $t \propto \frac{1}{\sqrt{\Delta l}}$ if $\Delta l \lt \lt l$
We can find an expression for $v$ if $\Delta l \gt \gt l$:
$v = \sqrt{\frac{k~\Delta l~(l+\Delta l)}{m}} \approx \sqrt{\frac{k~(\Delta l)^2}{m}}$
We can find an expression for the time if $\Delta l \gt \gt l$:
$t = \frac{l+\Delta l}{v}$
$t \approx \frac{\Delta l}{\sqrt{\frac{k~(\Delta l)^2}{m}}}$
$t \approx \sqrt{\frac{m}{k}}$
Therefore, $t$ is constant if $\Delta l \gt \gt l$
