Physics olympiad question about a charge capacitor
After giving it some thought, here is a new answer:
The answer from the book is incorrect. Your derivation is correct.
They probably made the same mistake as I did in my other answer, which I now believe is incorrect. The mistake I made is that I ignored the capacitance of the resistor.
I thought it would be easier to see the system as two capacitors with a resistor in between. However it only makes things more complicated. It is thus easier to follow your derivation using Gauss' law. Still, in this answer I will show that also when you consider the system to consist of two capacitors with a resistor in between, the answer you obtain differs from that in the book.
When a voltage is put over a resistor, there is a small surface charge at the interface between the leads and the resistor. This surface charge is given by:
$$\rho=\frac{\epsilon_0 U S}{L},$$
with $L$ the lenght of the resistor. This causes a resistor to act as a smalll capacitor. This paracitic capacitance and the corresponding charge are typically very small for common resistors. However, in this specific case, the resistor formed by the slab has a shape with $\sqrt{S}>>L$. This causes the resistor to have a large capacitance. The slab should therefore be represented as an ideal resistor with a capacitance placed parallel to it.
Our total system thus consists two capacitors formed by the gaps between the slab and the plate, one ideal resistor in between them and parallel to this resistor, the capacitance of the slab.
When the total system is charged there will be a charge
$$\pm q_{gap}=\pm \frac{\epsilon_0 U_0 S}{d-h}$$
on either side of the gap capacitors. The paracitic capacitor will remain uncharged. After shorting the leads, there initially will be an voltage drop over some of the wires. Since our wires are ideal, this will be compensated by charging the paracitic capacitance and slightly discharging the the gap capacitances. This happens instantaneously and no current flows through the resistor. This proces stops when there is no longer any voltage drop over the wires, i.e., when the voltage over the paracitic capacitance matches the voltage total voltage drop over the gap capacitances.
At that point the gap capacitances will have a charge of
$$\pm q'_{gap} = \frac{\epsilon_0 U_0 S}{d-h}\frac{h}{d}$$
while the paracitic capacitor has charges.
$$\pm q'_{para} = \pm \frac{\epsilon_0 U_0 S}{d-h}\left(1-\frac{h}{d}\right)=\pm \frac{\epsilon_0 U_0 S}{d} $$
Note that in reality the surface of the slab acts as a capacitor plate for both the gap capacitance and the paracitic capacitance. The total charge on this surface does not change during the initial charge transfer.
The voltage over the paracitic capacitance is equal to the voltage over the resistor and given by:
$$U_R = U_0\frac{h}{d}$$
this leads to
$$I_{max}=\frac{U_R}{R} = U_0 \frac{S}{\rho d}$$
This is the same answer you found using Gauss' law, which in this case is much more simple. The answer in the book is thus incorrect.