The four coupled differential equations $$\frac{dv_x}{dt}=-\frac{GM_{\odot}}{r^3}x,$$ $$\frac{dx}{dt}=v_x,$$ $$\frac{dv_y}{dt}=-\frac{GM_{\odot}}{r^3}y,$$ $$\frac{dy}{dt}=v_y,$$ can be turned into dimensionless equations (as we did in project 2) or we can introduce astronomical units with $1$ AU = $1.5\times 10^{11}$.

Using the equations from circular motion (with $r =1\mathrm{AU}$) $$\frac{M_E v^2}{r} = F = \frac{GM_{\odot}M_E}{r^2},$$ we have $$GM_{\odot}=v^2r,$$ and using that the velocity of Earth (assuming circular motion) is $v = 2\pi r/\mathrm{yr}=2\pi\mathrm{AU}/\mathrm{yr}$, we have $$GM_{\odot}= v^2r = 4\pi^2 \frac{(\mathrm{AU})^3}{\mathrm{yr}^2}.$$