In this case,
speed of belt w.r.t. ground $\therefore v_{B C}=4 \mathrm{\,km} \mathrm{h}^{-1}$
Speed of child wr.t. belt $\quad \therefore v_{C B}=9 \mathrm{\,km} \mathrm{h}^{-1}$
$\therefore$ For an observer on a stationary platform, speed of child running in the direction of motion of the belt is $\mathrm{v}_{\mathrm{CG}}=\mathrm{v}_{\mathrm{CB}}+\mathrm{v}_{\mathrm{BG}}=9 \mathrm{\,km} \mathrm{h}^{-1}+4 \mathrm{\,km} \mathrm{h}^{-1}=13 \mathrm{\,km} \mathrm{h}^{-1}$