Solve the Problem $\mathrm { S } ( \mathrm { r } ) = 4 \pi \mathrm { r } ^ { 2 }$

Solve the problem.
-The surface area of a balloon is given by $\mathrm { S } ( \mathrm { r } ) = 4 \pi \mathrm { r } ^ { 2 }$ , where $\mathrm { r }$ is the radius of the balloon. If the radius is increasing with time $t$ , as the balloon is being blown up, according to the formula $r ( t ) = \frac { 2 } { 3 } t { } ^ { 3 } , t \geq 0$ , find the surface area $S$ as a function of the time $t .$

A) $S ( r ( t ) ) = \frac { 16 } { 9 } \pi t ^ { 6 }$
B) $S ( r ( t ) ) = \frac { 16 } { 9 } \pi t ^ { 9 }$
C) $S ( r ( t ) ) = \frac { 4 } { 9 } \pi t ^ { 6 }$
D) $S ( r ( t ) ) = \frac { 16 } { 9 } \pi t ^ { 3 }$