Sphere

$V=\frac43\pi r^3$

$\dfrac {dV}{dt}=4\pi r^2\dfrac{dr}{dt}$

$V'(3)=4\pi \times9\dfrac{dr}{dt}=2$

$\dfrac{dr}{dt}=\dfrac{1}{18\pi}$

$S=4\pi r^2$

$\dfrac{dS}{dt}=8\pi r \dfrac{dr}{dt}$

$S'(3)=8\pi\times 3\times\dfrac{1}{18\pi}=\boxed{\dfrac43}$

${\text{Volume of sphere}} = \dfrac{4}{3} \pi r^3 \\ \dfrac{dV}{dt} = 4 \pi r^2 \cdot \dfrac{dr}{dt} \\ \implies 2 = 4 \pi r^2 \dfrac{dr}{dt} ...........................................(\because \dfrac{dV}{dt} {\text{is given}}) \\ \implies \dfrac{dr}{dt} = \dfrac{1}{2 \pi r^2} ...........................................(1) \\ {\text{Now, Surface area}} = 4 \pi r^2 \\ \implies \dfrac{dS}{dt} = 8 \pi r \cdot \dfrac{dr}{dt} \\ \implies \dfrac{dS}{dt} = \dfrac{8 \pi r}{2 \pi r^2}...........................................(by (1)) \\ \implies \dfrac{dS}{dt} = \dfrac{4}{r} \\ \dfrac{dS}{dt}_{r = 3 cm} = \color{#EC7300}{\boxed{\dfrac{4}{3} cm^2/sec}}$