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

Let the radius of the cone by r
Now, Volume cylindrical bucket = Volume of conical heap of sand

$\Rightarrow \pi(18)^{2}(32)=\frac{1}{3} \pi r^{2}(24)$

$\Rightarrow(18)^{2}(32)=8 r^{2}$

$\Rightarrow r^{2}=18 \times 18 \times 4$

$\Rightarrow r^{2}=1296$

$\Rightarrow r=36 \mathrm{~cm}$

Let the slant height of the cone be l.

Thus , the slant height is given by

$l=\sqrt{(24)^{2}+(36)^{2}}$

$=\sqrt{576+1296}$

$=\sqrt{1872}$

$=12 \sqrt{13} \mathrm{~cm}$

Disclaimer: The answer given in the book for the slant height is not correct.