Time constant of first order system
Set \$ t = \tau \$ in your equation. This gives

$$ y(\tau) = Ku(\tau)(1-e^{- \frac {\tau}{\tau}}) = Ku (1 - e^{-1}) = Ku(1 - 0.368) = 0.632Ku $$

where K is the DC gain, u(t) is the input signal, t is time, \$ \tau \$ is the time constant and y(t) is the output.

The time constant can be found where the curve is 63% of the way to the steady state output.

Easy-to-remember points are \$ \tau \$ @ 63%, \$ 3 \tau \$ @ 95\% and \$ 5 \tau \$ @ 99\%.

Your calculation for \$ \tau = \frac {3}{5} \$ appears to be based on the time the curve reaches \$ 5 \tau \$ but that is very difficult to pinpoint on the gentle slope of the curve. In the graphical solution below it looks more like 0.5 s rather than 0.6 s. (\$ 3 \cdot 63\% = 1.89 \$.)

Full answer



\$ \tau \$ = 0.5.