An irrigation tank in Gujarat is a cone with its point down…

Step 1. The link. The small cone of water is similar to the tank, so $\dfrac{r}{h} = \dfrac{3}{9}$ and $r = \dfrac{h}{3}$. Then $V = \tfrac{1}{3}\pi\left(\dfrac{h}{3}\right)^2 h = \dfrac{\pi h^3}{27}$.

Step 2. The volume. At $h = 6\ \text{m}$, $V = \dfrac{216\pi}{27} = 8\pi = 25.13\ \text{m}^3$.

Step 3. The depth. $\dfrac{dV}{dt} = \dfrac{\pi h^2}{9}\,\dfrac{dh}{dt}$. So $2 = 4\pi\,\dfrac{dh}{dt}$ and $\dfrac{dh}{dt} = \dfrac{1}{2\pi} = 0.159\ \text{m/min}$.

Step 4. The surface. $A = \dfrac{\pi h^2}{9}$, so $\dfrac{dA}{dt} = \dfrac{2\pi h}{9}\,\dfrac{dh}{dt} = \dfrac{12\pi}{9} \times \dfrac{1}{2\pi} = 0.667\ \text{m}^2/\text{min}$.