Step 1:

Let $r$ be the radius,$h$ be the height and $v$ be the volume of the sand-cone at any time $t$.

$v=\large\frac{1}{3}$$\pi r^2h$

It is given $r=6h$

Therefore $v=\large\frac{1}{3}$$\pi(36h^2)h$

$v=12\pi h^3$

Differentiating w.r.t $t$ on both sides we get,

$\large\frac{dv}{dt}$$=36\pi h^2\large\frac{dh}{dt}$

Step 2:

It is given $\large\frac{dv}{dt}$$=12cm^3/sec$

Therefore $12=36\pi h^2.\large\frac{dh}{dt}$

$\Rightarrow \large\frac{dh}{dt}=\frac{1}{3\pi h^2}$

Step 3:

When $h=4$

$\big(\large\frac{dh}{dt}\big)_{at\;h=4}=\frac{1}{3\pi(4)^2}$

$\Rightarrow \large\frac{1}{48\pi}$

Hence the height of the sand cone is increasing at the rate of $\large\frac{1}{48\pi}$$cm/sec$