Interpret the object

A hollow ice-cream cone means a thin conical shell (not a solid cone), open at the top.

We need the moment of inertia about its symmetry axis.

Take a thin ring element

Consider a small circular ring on the conical surface at radius $r$ and slant distance $l$ from the vertex.

For a thin conical shell, surface mass density is uniform: $\sigma = \frac{M}{\pi R L}$ where $L = \sqrt{R^2 + H^2}$ is the slant height.

The area of a thin ring of width $dl$ is $dA = 2\pi r\, dl$

so its mass is $dm = \sigma \, dA = \sigma (2\pi r\, dl)$

Relation between $r$ and $l$

By similarity in the cone, $\frac{r}{l} = \frac{R}{L}$ so $r = \frac{R}{L}l$

Moment of inertia of the ring

For a ring about its axis, $dI = r^2\, dm$

Substitute $dm$: $dI = r^2 \cdot \sigma (2\pi r\, dl) = 2\pi \sigma r^3 dl$

Using $r = \frac{R}{L}l$,

Integrate from vertex to rim

$I = \int_0^L dI = 2\pi \sigma \frac{R^3}{L^3} \int_0^L l^3 dl$

$I = 2\pi \sigma \frac{R^3}{L^3} \cdot \frac{L^4}{4}$

$I = \frac{\pi \sigma R^3 L}{2}$

Substitute $\sigma$

Since $\sigma = \frac{M}{\pi R L}$

we get $I = \frac{\pi}{2} \cdot \frac{M}{\pi R L} \cdot R^3 L$

$I = \frac{MR^2}{2}$

Match with options

$I = \frac{MR^2}{2}$

So the correct option is B.

Comparison with stored answer

Stored correct answer: B

Derived answer: B

They agree.