As per the question,

The diameter of the cylinder (also the same for cone) $=24m$.

Thus, its radius (R) $=24/2=12m$

The height of the cylindrical part $\left( {{H}_{1}} \right)=11m$

Now, Height of the cone part $\left( {{H}_{2}} \right)=16-11=5m$

Then,

Vertex of the cone above the ground $=11+5=16m$

Curved Surface area of the Cone $\left( {{S}_{1}} \right)=\pi rl=22/7\times 12\times L$

Give the slant height (L) is,

$L=\sqrt{\left( {{R}^{2}}+H_{2}^{2} \right)}=\sqrt{\left( {{12}^{2}}+{{5}^{2}} \right)}=\sqrt{169}$

$L=13m$

Thus,

Curved Surface Area of Cone$\left( {{S}_{1}} \right)=22/7\times 12\times 13$

Then,

Curved Surface Area of Cylinder $\left( {{S}_{2}} \right)=2\pi R{{H}_{1}}$

${{S}_{2}}=2\pi \left( 12 \right)\left( 11 \right){{m}^{2}}$

Thus, the area of Canvas required for tent

$S={{S}_{1}}+{{S}_{2}}=\left( 22/7\times 12\times 13 \right)+\left( 2\times 22/7\times 12\times 11 \right)$

$S=490+829.38$

$S=1319.8{{m}^{2}}$

$S=1320{{m}^{2}}$

Hence, the area of canvas required for the tent is $1320{{m}^{2}}$.