Let $h(t)$ be the height of the water at time $t$, and let $r(t)$ be the radius of the surface of the water at time $t$; from similar triangles we know that
$$\frac{h(t)}{r(t)}=\frac{14}{2.75}=\frac{56}{11}\;.$$
We could solve this for either $r(t)$ or $h(t)$ in terms of the other, but notice that we’re told $h'(t)$ at a particular moment, and we’re not told anything about an specific value of $r(t)$. This suggests that we’d be better off working in terms of $h(t)$, so we’ll solve for $r(t)$ and get $$r(t)=\frac{11}{56}h(t)\;.$$
At time $t$ the volume $V(t)$ of water in the tank is the volume of a right circular cone with height $h(t)$ and base radius $r(t)$, which is given by
$$V(t)=\frac13\pi r(t)^2h(t)=\frac{\pi}3\left(\frac{11}{56}h(t)\right)^2h(t)=\frac{121\pi}{9408}h(t)^3\;.$$
Then
$$V'(t)=\frac{121\pi}{3136}h(t)^2h'(t)\;.$$
We’re told that $h'(t)=0.24$ when $h(t)=3$; if we call that moment time $t_0$, we have
$$V'(t_0)=\frac{121\pi}{3136}\cdot3^2\cdot0.24=\frac{3267\pi}{39200}\text{ m}^3/\text{min}\;.$$
Now let $v$ be the rate in cubic metres per minute at which water is being pumped into the tank. Taking into account both the inflow and the leakage, we know that at all times
$$V'(t)=v-0.0068\text{ m}^3/\text{min}\;.$$
In particular, at time $t_0$ we have
$$v-0.0068=\frac{3267\pi}{39200}\;,$$
which is completely straightforward to solve for $v$.
Think about what is happening. The water (volume) is being poured in at a constant rate. This relates to how the water level ($h$) changes and how the width of the water in the tank at that level ($r$) changes. Further, $r$ and $h$ are related.
The volume of a cone is
$$V = \frac13 \pi \, r^2 \, h$$
How is $h$ related to $r$? You know that, at the top, the radius is $2$ and $h=4$. Because this is a cone, we can say that $r = h/2$ at all levels of the cone. Thus,
$$V(h) = \frac{1}{12} \pi \, h^3$$
We may then differentiate with respect to time; use the chain rule here
$$\frac{dV}{dt} = \frac{\pi}{4} h^2 \frac{dh}{dt}$$
You are given $dV/dt$ and the height $h$ at which to evaluate; solve for $dh/dt$.
Best Answer
The change in the volume of the water is the difference in the fill and leak rates (taking the leak rate to be a positive number).
$$\frac{dV}{dt} = \frac{dF}{dt} - \frac{dL}{dt}$$ Thus, the leak rate is the difference in the fill rate and the actual rate of increase in the volume of the water. $$\frac{dL}{dt} = \frac{dF}{dt} - \frac{dV}{dt}$$ The volume of the water when its height is $h$ and its radius is $r$ is $$V = \frac{1}{3}\pi r^2h$$ Since the inverted conical tank has a diameter of $10~\text{ft}$ at the top, its radius at the top is $5~\text{ft}$.
By similar triangles, the ratio of the radius of the water to the height of the water in the cone is $$\frac{r}{h} = \frac{5~\text{ft}}{20~\text{ft}} = \frac{1}{4} \implies r = \frac{h}{4}$$ Thus, we can express the volume of the water as a function of $h$ by substituting $h/4$ for $r$, which yields $$V(h) = \frac{1}{3}\pi\left(\frac{h}{4}\right)^2h = \frac{1}{48}\pi h^3$$ Differentiating with respect to time yields $$\frac{dV}{dt} = \frac{1}{16}\pi h^2 \frac{dh}{dt}$$ We are given that $$\frac{dh}{dt} = 1~\frac{\text{in}}{\text{min}} = \frac{1}{12}~\frac{\text{ft}}{\text{min}}$$ when $h = 16~\text{ft}$. Thus, the rate at which the volume is increasing is $$\frac{dV}{dt} = \frac{1}{16}\pi (16~\text{ft})^2\left(\frac{1}{12}~\frac{\text{ft}}{\text{min}}\right) = \frac{4}{3}\pi~\frac{\text{ft}^3}{\text{min}}$$
Since the fill rate is $8~\text{ft}^3/\text{min}$, the leak rate is $$\frac{dL}{dt} = \frac{dF}{dt} - \frac{dV}{dt} = 8~\frac{\text{ft}^3}{\text{min}} - \frac{4}{3}\pi~\frac{\text{ft}^3}{\text{min}} = \left(8 - \frac{4}{3}\pi\right)~\frac{\text{ft}^3}{\text{min}}$$