### Solution:

Given,Radius of cylindrical container, r = 6cm

Height of cylindrical container, h = 15cm

Height of the conical portion, H = 4 × radius of base of cone, x

Let the radius of the ice-cream cone be x, then the shape of an ice-cream cone with hemispherical top is given below.

According to question,

Volume of cylindrical container = 10 × volume of ice cream cone

Also,

Volume of ice-cream cone = volume of hemispherical top + volume of cone

Now,

We know that the volume of a cylinder of radius, r and height, h is given by $\n \n \pi \n r\n 2\n \n h\n $ .

We get,

$\n \n \n \n \n V\n \n cylinder\n \n =\pi \n r\n 2\n \n h\n \n \n \n \n \u21d2\n V\n \n cylinder\n \n =\pi \n (6)\n 2\n \n 15\n \n \n \n \n $

And

The volume of a cone of radius, r and height, h is given by $\n \n \n 1\n 3\n \n \pi \n r\n 2\n \n h\n $ .

The volume of a hemisphere of radius, r is given by $\n \n \n 2\n 3\n \n \pi \n r\n 2\n \n h\n $ . $\n \n Volumeofoneice-creamcone=\n 2\n 3\n \n \pi \n r\n 3\n \n +\n 1\n 3\n \n \pi \n r\n 2\n \n h\n $

We get,

$\n \n \n \n Volumeofoneice-creamcone=\n 2\n 3\n \n \pi \n r\n 3\n \n +\n 1\n 3\n \n \pi \n r\n 2\n \n h\n \n \n \n \n \u21d2Volumeofoneice-creamcone=\n 2\n 3\n \n \pi \n x\n 3\n \n +\n 1\n 3\n \n \pi \n x\n 2\n \n \n \n 4x\n \n \n \n \n \n \u21d2Volumeofoneice-creamcone=2\pi \n x\n 3\n \n \n \n \n \n \n $

Now,

Volume of cylindrical container = 10 × volume of ice cream cone

$\n \n \n \n \pi \n (6)\n 2\n \n 15=10\xd72\pi \n x\n 3\n \n \n \n \n \n \n \u21d2\n x\n 3\n \n =\n \n \pi \n \n (6)\n 2\n \n 15\n \n 10\xd72\pi \n \n \n \n \n \n \n \u21d2\n x\n 3\n \n =\n \n 6\xd76\xd715\n \n 10\xd72\n \n \n \n \n \n \n \u21d2\n x\n 3\n \n =3\xd73\xd73\n \n \n \n \n \u21d2x=3cm\n \n \n \n \n $

Therefore, the radius of the ice-cream cone is 3 cm.

Hence, the correct option is 1.