Answer
$3.0 min$
Work Step by Step
We know that
$\Delta t_1=\frac{\Delta t_{\circ}}{\sqrt{1-v_1^2/c^2}}$
$\implies \Delta t_{\circ}=\Delta t_1\sqrt{1-v_1^2/c^2}$
Now we can find $\Delta t_2 $ as
$\Delta t_2=\frac{\Delta t_{\circ}}{\sqrt{1-v^2/c^2}}=\frac{
\Delta t_1\sqrt{1-v^2/c^2}}{\sqrt{1-v^2/c^2}}$
We plug in the known values to obtain:
$\Delta t_2=(5.0min)\sqrt{\frac{1-(0.95)^2}{1-(0.80)^2}}=3.0 min$
