Let $y$ be the depth of the water in this cone. Then $y$ is a function of time $t$. The target of this problem is to compute $\frac{y}{t}$ when $y=\frac{H}{2}$. 

It is clear that $\frac{y}{t}$ is related to the change in the volume of accumulated water $V$. Because the value of $V$ is related $y$ by (the difference of two cones)

$$V=\frac{1}{3}\pi R^2H - \frac{1}{3}\pi \left(\frac{R(H-y)}{H}\right)^2(H-y)=\frac{\pi R^2}{3H^2}\left(H^3-(H-y)^3\right)$$

the change of volume at the moment $t$ can be computed as (note that $y$ is a function of $t$, thus we must apply the chain rule)

$$\frac{V}{t}=\frac{\pi R^2}{H^2}(H-y)^2\frac{y}{t}$$

Apparently, the change in the volume of accumulated water must equal to the speed of which water is poured in, i.e. $\frac{V}{t}=A$. Setting this to the relation above yields

$$\frac{y}{t}=\frac{AH^2}{\pi R^2(H-y)^2}$$

Finally, we find when $y=\frac{H}{2}$, $$\frac{y}{t}=\boxed{\frac{4A}{\pi R^2}}$$