How do you find the rate at which water is pumped into an inverted conical tank that has a height of 6m and a diameter of 4m if water is leaking out at the rate of #10,000(cm)^3/min# and the water level is rising #20 (cm)/min#?

This question has already been answered although you seem to be missing the height of the water in the cone at the time the water level is rising #20 (cm)/(min)#.
Assuming this question came from the same source, the specified height of water was #2 m# or #200 cm#.

The cone has a radius of 2 m (half the diameter) and a height of 6 m
for a ratio of #(radius)/(height) = 1/3#

This ratio is constant for volumes of water contained in the cone,

Therefore the volume of the cone (or water in the cone), normally written as
#V(r,h) = (pi r^2h)/3#

can be re-written as
#V(h) = ( pi (h/3)^2 h)/3#
#= (pi h^3)/(27)#

and therefore
#(d V(h))/(dh) = pi/9 h^2# #(cm^3)/(cm)#

We are told
#(d h)/(dt) = 20 (cm)/(min)#

The increase in volume contained in the cone is given by
#(d V)/(dh) xx (d h)/(dt)# at water level height of #200 cm#

#= pi/9 (200 cm)^2 xx 20 (cm)/(min)#
#= 2,792,527 (cm^3)/(min)# (approx. assuming I haven't slipped up somewhere)

The inflow of water must be the total of
the outflow (leakage)
plus
the amount needed to raise the water level:
#10,000 ((cm)^3)/(min)#
#+ 2,792,527 ((cm)^3)/(min)#

#= 2,802,527 ((cm)^3/min)#