Answer: Option C
$$\eqalign{
& h = 14\,cm,\,r = 7\,cm \cr
& {\text{So}},\,l = \sqrt {{{\left( 7 \right)}^2} + {{\left( {14} \right)}^2}} = \sqrt {245} = 7\sqrt 5 \,cm \cr
& \therefore {\text{Total}}\,{\text{surface}}\,{\text{area}} \cr
& = \pi \,rl + \pi \,{r^2} \cr
& = \left( {\frac{{22}}{7} \times 7 \times 7\sqrt 5 + \frac{{22}}{7} \times 7 \times 7} \right)c{m^2} \cr
& = \left[ {154\left( {\sqrt 5 + 1} \right)} \right]c{m^2} \cr
& = \left( {154 \times 3.236} \right)c{m^2} \cr
& = 498.35\,c{m^2} \cr} $$