Answer
$t = 1.8 s$
Work Step by Step
In an $RC$ circuit charge $q$ gained in time $t$ is given as
$ q= q_0[1- e^{-t/(RC)} ]$
but $q= \frac{q_0}{2}$ and $\tau =RC =2.6 s$
$\therefore \frac{q_0}{2} = q_0[1- e^{-t/(RC)} ] $
$ e^{-t/(2.6s)} = \frac{1}{2}$
Taking log on both sides
$ \frac{-t}{2.6s} =ln (\frac{1}{2})$
$ t = 2.6s * ln 2$
$t = 1.8 s$
