In this case, $u=\sqrt{x^2+y^2}$ and $\Omega=\{x^2+y^2 \le 2020\}.$ So, if $$u=\sqrt{x^2+y^2} \Rightarrow u_x=\frac{x}{\sqrt{x^2+y^2}} \Rightarrow u_{xx}=\frac{\sqrt{x^2+y^2}-x^2(x^2+y^2)^{-\frac{3}{2}}}{x^2+y^2}$$
$$u_y=\frac{y}{\sqrt{x^2+y^2}}\Rightarrow u_{yy}=\frac{\sqrt{x^2+y^2}-y^2(x^2+y^2)^{-\frac{3}{2}}}{x^2+y^2}$$ Then $$\int_{\Omega} \frac{2\sqrt{x^2+y^2}-(x^2+y^2)(x^2+y^2)^{-\frac{3}{2}}}{x^2+y^2} \, dx \, dy = \\ 2\int_\Omega \frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2(x^2+y^2)}} \, dx \, dy = \\ 2\int_{\Omega}\frac{1}{x^2+y^2} \, dx \, dy=\left[r^2=x^2+y^2 \right] \Rightarrow \\ 2\int_{\Omega}\frac{r}{r^2} \, dr \, d\phi$$
I think this is incorrect and I don't know how to proceed further. It is given that $x^2+y^2 \le 2020$, so maybe I can take bounds for $r$ from 0 to 2020? Don't know about $\phi$ either.
You messed up right away in that $$\begin{align}u_{xx}&=\frac{\partial}{\partial x}\left(x(x^2+y^2)^{-1/2}\right)=(1)(x^2+y^2)^{-1/2}+(x)\left(-\frac12\right)(x^2+y^2)^{-3/2}(2x)\\ &=\frac{x^2+y^2-x^2}{(x^2+y^2)^{3/2}}=\frac{y^2}{(x^2+y^2)^{3/2}}\end{align}$$ It should be anticipated that $$u_{yy}=\frac{x^2}{(x^2+y^2)^{3/2}}$$ So that $$\nabla^2u=u_{xx}+u_{yy}=\frac{x^2+y^2}{(x^2+y^2)^{3/2}}=\frac1{\sqrt{x^2+y^2}}$$ You could have gotten this also from $$\nabla^2f(r)=\frac1r\frac{\partial}{\partial r}\left(r\frac{\partial f(r)}{\partial r}\right)$$ In $2$ dimensions. Then $$\int\int_{\Omega}\nabla^2u\,d^2A=\int_0^{2\pi}\int_0^{\sqrt{2020}}\frac1r\cdot r\,dr\,d\theta=2\pi\sqrt{2020}$$ On $\partial\Omega$, $$\vec r=\sqrt{2020}\langle\cos\theta,\sin\theta\rangle$$, $$d\vec r=\sqrt{2020}\langle-\sin\theta,\cos\theta\rangle\,d\theta$$ So $$\hat n\,d\ell=\sqrt{2020}\langle-\sin\theta,\cos\theta,0\rangle\,d\theta\times\langle0,0,1\rangle=\sqrt{2020}\langle\cos\theta,\sin\theta,0\rangle\,d\theta$$ Also $$\vec\nabla u=\langle\cos\theta,\sin\theta,0\rangle$$ So $$\oint_{\partial\Omega}\vec\nabla u\cdot\hat n\,d\ell=\int_0^{2\pi}\langle\cos\theta,\sin\theta,0\rangle\cdot\sqrt{2020}\langle\cos\theta,\sin\theta,0\rangle\,d\theta=2\pi\sqrt{2020}$$