According to the given information,
Let us asssume the radius of the small cone be r cm
And, the radius of the big cone be R cm
It is given, height of the big cone is $20cm$
Let us also assume the height of section made be h cm
Therefore, the height of small cone will be $(20–h)cm$
Now according to the question,
In $\vartriangle OAB$ and $\vartriangle OCD$
$\angle AOB=\angle COD$ [common]
$\angle OAB=\angle OCD$ [each ${{90}^{\circ }}$]
Therefore, $\vartriangle OAB\sim \vartriangle OCD$ [by AA similarity]
So, by corresponding part of similar triangles we have,
$OA/OC=AB/CD$
By subsituiting their value,
$(20–h)/20=r/R$ …… (i)
It is also given in the question that,
Volume of small cone $=1/125\times $ volume of big cone
$1/3\pi {{r}^{2}}\left( 20-h \right)=1/125\times 1/3\pi {{R}^{2}}\times 20$
${{r}^{2}}/{{R}^{2}}=1/125\times 20/\left( 20-h \right)$ [From (i)]
${{\left( 20-h \right)}^{2}}/{{20}^{2}}=1/125\times 20/20-h$
${{\left( 20-h \right)}^{2}}={{20}^{3}}/125$
$20-h=20/5$
$20-h=4$
$h=20-4$
$=16cm$
Therefore, 16cm above the base was the height of the section made.