Answer: Option B
$$\eqalign{
& \pi {r^2} = 346.5 \cr
& {r^2} = \left( {346.5 \times \frac{7}{{22}}} \right) \cr
& {r^2} = \frac{{441}}{4} \cr
& {r^2} = \frac{21}{2} \cr
& \therefore l = \sqrt {{r^2} + {h^2}} \cr
& \,\,\,\,\,\,\, = \sqrt {\frac{{441}}{4} + {{\left( {14} \right)}^2}} \cr
& \,\,\,\,\,\,\, = \sqrt {\frac{{1225}}{4}} \cr
& \,\,\,\,\,\,\, = \frac{{35}}{2} \cr} $$
So, area of canvas needed :
$$\eqalign{
& \pi rl = \left( {\frac{{22}}{7} \times \frac{{21}}{2} \times \frac{{35}}{2}} \right){{\text{m}}^2} \cr
& \,\,\,\,\,\,\,\,\,\,\, = \left( {\frac{{33 \times 35}}{2}} \right){{\text{m}}^2} \cr} $$
∴ Length of canvas :
$$\eqalign{
& = \left( {\frac{{33 \times 35}}{{2 \times 1.1}}} \right){\text{ m}} \cr
& = 525{\text{ m}} \cr} $$