moment of Inertia of the base section about a diameter is

$ I = {\pi\ times(D^4-d^4)\over 64 }={\pi\times(3^4-2^4) \over 64 }= {3.191 m^4 }$

area of the base is given by
${\pi\times (D^2-d^2)\over2}={\pi\times (3^2-2^2)\over 4}={3.927 m^2}$

section of the base $Z = {2I\over D }= {2\times3.191\over 3}={2.127}$

direct stress due to the weight of the chimney ${1800000\over 3.927}={458365.16}$

AREA OF THE CHIMNEY IS GIVEN BY THE FOLLOWING FORMULAE

${h\times (D+d)\over 2}= {62.5}$

If p is the wind pressure of the shear force P then it acts at the centroid of the structure then
${P }= {62.5\times p}$ is the total wind pressure
centroid of the trapezium is given by the following formula $\bar{y}= {(D+ 2\times d )\over(D+d)}{\times h \over 2}= {(3+ 2\times2)\times h\over 3+2 }= 11.65$
moment due to the pressure

$M =P\times \bar{y}$

${P{b}} ={\pm\ M\over Z}= {62.5 \times p\times11.67\over2.127}$
Pb =Po
$ {62.5 \times p\times11.67\over2.127}$ =48365

so p =1336.87 $N\over m^2 $