The amplitude of $\displaystyle{y}={a}{\sin{{b}}}{x}\usepackage{cmbright}\Huge\$ is given by $\displaystyle{\left|{a}\right|}\usepackage{cmbright}\Huge\$ , so the amplitude of $\displaystyle{y}={2}{\sin{{\left(\frac{{x}}{{3}}\right)}}}\usepackage{cmbright}\Huge\$ is 2. The period equals $\displaystyle\frac{{{2}\pi}}{{\left|{b}\right|}}\usepackage{cmbright}\Huge\$ . Now $\displaystyle{2}{\sin{{\left(\frac{{x}}{{3}}\right)}}}={2}{\sin{{\left(\frac{{1}}{{3}}{x}\right)}}}\usepackage{cmbright}\Huge\$ , so $\displaystyle{b}=\frac{{1}}{{3}}\usepackage{cmbright}\Huge\$ and the period equals:

$\displaystyle\frac{{{2}\pi}}{{\left|\frac{{1}}{{3}}\right|}}=\frac{{{2}\pi}}{{\frac{{1}}{{3}}}}=\frac{{{2}\pi}}{{1}}\cdot\frac{{3}}{{1}}={6}\pi\usepackage{cmbright}\Huge\$