Therefore, $P=\rho g h$

Let's consider the topmost point as the base reference to be at zero pressure.

As we move below, consider a semi-circular ring of radius $r$ and thickness $dr.$

The height of this ring from the topmost reference point $=\mathrm{R}-\mathrm{r}$

Hence, the pressure outside any such ring, $\mathrm{dP}=\rho g(R-r)$

The force exerted by the fluid on any such ring, $\mathrm{dF}=d P \times A$

Hence, dF $=\rho g(R-r) \times 2 \pi r d r$

The counteracting force required to keep the gate stable is the integration of this pressure exerted force.

$\mathrm{F}=\int_{0}^{R} \rho g(R-r) \times 2 \pi r d r$

$F=\frac{\pi \rho g R^{3}}{3}$

Hence, the answer is none of these.