This problem seems best treated in cylindrical coordinates. There is azimuthal symmetry, and there is no heat loss or generation within the cone, so our thermal conductivity equation reads:
$$\vec{q} = -k(\frac{\partial T}{\partial \rho} \hat{\rho} + \frac{\partial T}{\partial z} \hat{z})$$
We...
We can write our radius as a function of the height, z, of our cone: $$R(z) = \frac{R_2 - R_1}{h} z + R_1$$
Where h is the height of our cone, ##h = \frac{L}{40}##.
Our cross sectional area, $$A = 2 \pi R t$$ can then be written as $$A = 2 \pi t [\frac{R_2 - R_1}{h} z + R_1]$$
This I am all...