By "inverted conical tank", I'm going to assume the pointy end is pointed downwards. This only mildly affects how the problem is solved.
Let's give some names to the values we're working with:
t = time in minutes
V(t) = Volume of water in cm^3 in tank at t minutes
h(t) = height of water in the tank in cm (starting from pointy end) at t minutes
C = constant rate in cm^3/min of water flowing in that we're looking for
We can tell from the outset that this is a related rates problem since it discusses the rate of change of two related values: the volume of water in the tank and the height of the water in the tank. Thus, we should try to find the relationship between these two values.
If you can imagine a cone with the pointy end down, the volume of water in this cone in relation to the height of the water inside it is simply the volume of a cone with height h (if you cut off the top of the cone at height h, you just get a cone of height h) and some radius. To find the radius r of the new cone, notice that taking a vertical cross-section reveals two similar right triangles: one with legs h and r, and one with legs 15m and 3.5m, respectively. We thus obtain the relation
h/15 = r/3.5 --> r = 3.5h/15 = 7h/30.
So the volume of water in the tank given the height of the water is V = π(h/3)(7h/30)^2, applying the formula for the volume of a cone.
If water is flowing in at a rate of C cm^3/min and flowing out at 8200 cm^3/min, then the total water flowing into the tank is C - 8200 cm^3/min. In particular, dV/dt = C - 8200 cm^3/min, since the derivative of the volume is the rate of change of that volume.
Now it all comes together. To find C, we're going to use the volume equation we obtained and take V and h to be functions of t. Carefully taking the derivative of both sides of V = π(h/3)(7h/30)^2 with respect to t gives
dV/dt = π(dh/dt)(7h/30)^2.
So after substituting dV/dt = C - 8200 cm^3/min, dh/dt = 20 cm/min, and h = 4m = 400cm, we obtain
C - 8200 cm^3/min = π(20 cm/min)(2800/30 cm)^2
or C ≈ 555535.25 cm^3/min.
As a note, if the cone were facing base-side down, our volume equation would instead be the total volume of the cone minus the area of a cone with height 1500 - h cm and corresponding radius.
