Let's write the formula for the volume of water in the tank as a function of depth $h$ and length $l$:
$$V = hl^2.$$
Now, we have that both volume $V$ and depth $h$ are functions of time:
$$V(t) = h(t)l^2.$$
We know that $dV/dt = 0.2 m^3/hr$. We also know that
$$\frac{dV(t)}{dt} = \frac{d}{dt}\left(h(t)l^2\right) = l^2\frac{dh(t)}{dt}.$$
It should be straightforward from here.
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 $v$.
Best Answer
There are numerous problems of this kind on this site copied from various exercises in homework or textbooks, describing a tank of some particular shape, saying that water enters the tank at a certain rate of volume per unit time, and asking how fast the water level in the tank is increasing when the water is at a certain level in the tank.
The "obvious" way to work these problems is to compute the volume of the water as a function of the level it has reached, then differentiate this volume relative to the variable that tells the level of the water. Using this derivative, $dV/dh,$ and the given inflow of water, $dV/dt,$ you can easily compute $dh/dt,$ which is what you were asked to compute.
The trick is to understand that in any problem of this kind, $$V(h) = \int_{h_0}^h A(\eta) d\eta,$$ where $V(h)$ is the volume of water when the level is $h,$ where $A(\eta)$ is the surface area of the top of the water when the level is $\eta,$ and where $h_0$ is the level at the bottom of the tank. (We usually set things up so that $h_0 = 0,$ but as I'm about to show, this is generally irrelevant.)
Applying the fundamental theorem of calculus, we can conclude that $$\frac{d}{dh} V(h) = A(h).$$ It's often harder to compute $V(h)$ than $A(h)$--in this case you have to compute $A(h)$ and then integrate it in order to get $V(h)$--and on top of that the "obvious" method then forces you to differentiate $V(h)$. So you integrate $A(h)$ and then differentiate the result, getting back $A(h)$; what a waste of effort!
In the easy approach, we just use the fact that $dV/dh = A(h),$ compute $A(h),$ and get a simple answer like Phil H's.