The diagram is faulty. Switch S1 needs to be in the same branch as the capacitor to prevent it from discharging before S1 is closed. But we can guess what they mean.
Using KVL is not much more difficult than what you have done already. It should give you the correct answer - for example, as follows.
Mark 3 loop currents $i_1, i_2, i_3$ as shown in the diagram below. Then applying KVL around each loop we get $$\frac{Q}{C}=3Ri_1-Ri_2$$ $$-\frac{Q}{C}=4Ri_2-Ri_1-2Ri_3$$ $$E=4Ri_3-2Ri_2$$ We are interested in the current through the capacitor which is $i=i_1-i_2$. Substituting and eliminating $i_1, i_2, i_3$ we obtain
$$i=-\frac{dQ}{dt}=\frac{1}{2CR}(Q-\frac14 CE)$$
$$Q-\frac14 CE=(Q_0-\frac14 CE)e^{-t/2RC}$$
After a long time ($t \to \infty$) the factor $e^{-t/2RC} \to 0$. Then the charge on the capacitor becomes $Q_{\infty}=\frac14CE$.
A quicker solution is to find the equivalent resistance $R'$ across the capacitor after shorting the cell. The time constant of the decay (or increase) of charge on the capacitor is then $C R'$. If the initial charge is $Q_0$ and the final charge is $Q_1$ then the time-variation of the charge is given by
$$ Q(t) - Q_1 = (Q_0 - Q_1) e^{-t / C R' }$$
After shorting the cell the circuit can be drawn as on the right above, because B and C are at the same potential. The resistance between C and D is $R$ because this is the result of two resistors $2R$ in parallel. Then the resistance along ADC is $2R$. This is the same as the resistance along ABC. ADC and ABC are in parallel, and have an equivalent resistance $R$. This is in series with the resistor in branch AC. So the effective resistance in series with the capacitor is $R'=2R$. The time constant for the circuit is $CR'=2CR$.
