A cylindrical bucket, 32 cm high and 18 cm of radius of the base,

The height and radius of the cylindrical bucket are $h=32 \mathrm{~cm}$ and $r=18 \mathrm{~cm}$ respectively. Therefore, the volume of the cylindrical bucket is

$V=\pi r^{2} h$

$=\frac{22}{7} \times(18)^{2} \times 32$

The bucket is full of sand and is emptied in the ground to form a conical heap of sand of height $h_{1}=24 \mathrm{~cm}$. Let, the radius and slant height of the conical heap be $r_{1} \mathrm{~cm}$ and $l_{1} \mathrm{~cm}$ respectively. Then, we have

$l_{1}^{2}=r_{1}^{2}+h_{1}^{2}$

$\Rightarrow r_{1}^{2}=l_{1}^{2}-h_{1}^{2}$

$\Rightarrow r_{1}^{2}=l_{1}^{2}-(24)^{2}$

The volume of the conical heap is

$V_{1}=\frac{1}{3} \pi r_{1}^{2} h_{1}$

$=\frac{1}{3} \times \frac{22}{7} \times r_{1}^{2} \times 24$

$=\frac{22}{7} \times r_{1}^{2} \times 8$

Since, the volume of the cylindrical bucket and conical hear are same, we have

$V_{1}=V$

$\Rightarrow \frac{22}{7} \times r_{1}^{2} \times 8=\frac{22}{7} \times(18)^{2} \times 32$

$\Rightarrow \quad r_{1}^{2}=(18)^{2} \times 4$

$\Rightarrow \quad r_{1}=18 \times 2$

$\Rightarrow \quad r_{1}=36$

Then, we have

$l_{1}^{2}=r_{1}^{2}+h_{1}^{2}$

$\Rightarrow l_{1}^{2}=(36)^{2}+(24)^{2}$

$\Rightarrow l_{1}=43.27$

Therefore, the radius and the slant height of the conical heap are 36 cm and 43.27 cm respectively.