Let the volume of water remaining in the container be
V(t), the flux at the leak be f(V), a function of the remaining water volume only.
dV/dt=-f(v).
The time it takes to drain the water is
T=\int dV/f(V), integrated from V=0 to V=V0.
In the two considered cases, f(V) with the wood is greater than f(V) without the wood, at any V. So the time with the wood is smaller.