You have forgotten that there is work done by the battery after $t=0$, so that there is energy lost before S1 is opened and S2 is closed.
The work done by the battery to transfer charge $Q$ onto the 1st capacitor is $W=QV=\frac{Q^2}{C}$.
The final charges on the capacitors are $\frac12Q, \frac14Q, \frac{1}{16}Q, \frac{1}{64}Q, ...$ The total energy stored in all the capacitors at this time is $$E_{\infty}=\frac{Q^2}{2C}[(\frac12)^2+(\frac14)^2+(\frac{1}{16})^2+(\frac{1}{64})^2 +...]=\frac{Q^2}{6C}$$ The total loss of energy after closing S1 is $$W-E_{\infty}=\frac{5Q^2}{6C}=\frac56CV^2=\frac{20}{6}\pi\epsilon_0 RV^2$$
So I think option A should be the correct answer. The marking scheme appears to be wrong.