Differentiation

I don't know why I'm wrong when I was answering a Q.
Please help me!

The question is:

A container in a form of inverted right circular cone, radius 6 cm and height 8 cm is initially full of water and then water is leaking away from the apex at a constant rate of 24 cm^3s^-1. Find the rate of change of the wetted curved surface area of the inner wall of the container when the depth of the water is 4 cm.

This is my answer:

Let V = volume of water.
Let t = time.
Let r = radius of the water surface.
Let h = height of the water.
Let l = slant height of the water.
Let A = wetted surface area of the container.

dV/dt = - 24
dV = - 24 dt --- (i)

V = 4/3 π r^3
dV/dr = 4 π r^2
dr = dV/(4 π r^2) --- (ii)

A = π r l = π r sqrt(r^2 + h^2) = π r sqrt[r^2 + (8/6 r)^2] = 5/3 π r^2
dA/dr = 10/3 π r
Sub (ii).
dA/[dV/(4 π r^2)] = 10/3 π r
dA/dV = 10/3 π r/(4 π r^2) = 10/(12 r)
Sub (i).
dA/(- 24 dt) = 10/(12 r)
dA/dt = - 20/r

When h = 4, r = 4(6/8) = 3.
dA/dt = - 20/3 cm^3 s^-1

But the answer should be 20 cm^3 s^-1

WHAT'S WRONG WITH MY ANSWER?

THANK YOU!~