Answer
$y=3\sin{\left(\frac{2}{3}x+\frac{2}{9}\right)}$.
Work Step by Step
The general form for the sinusoidal function is: $y=A\sin{(\omega x-\phi)}+B$ where $A$ is the amplitude, $B$ is the vertical shift, we need $\omega$ can be computed from the period by the formula $\omega=\frac{2\pi}{T}.$
The phase shift is $\frac{\phi}{\omega}$, hence $\phi=\omega\cdot\text{phase shift}.$
Hence here: $A=3$,
$B=0$,
$\omega=\frac{2\pi}{T}=\frac{2\pi}{3\pi}=\frac{2}{3}$
$\phi=\omega\cdot\text{phase shift}=\frac{2}{3}\cdot\frac{-1}{3}=\frac{-2}{9}$.
Hence our function is: $y=3\sin{\left(\frac{2}{3}x+\frac{2}{9}\right)}$.
