$G_2(\tau)=\mathop{\sum\limits_{c\in\mathbb{Z}}\sum\limits_{d\in\mathbb{Z}}}\limits_{(c,d)\neq(0,0)}\frac1{(c\tau+d)^2}=2\zeta(2)-8\pi^2\sum\limits_{n=1}^\infty\sigma(n)e^{2\pi i\tau n},\ \tau\in\mathcal{H}$.
It's easy to calculate that $G_2(i)=\pi$ and $G_2(\omega)=G_2(e^{\frac{2\pi i}3})=\frac{2\pi}{\sqrt{3}}$. But I can't calculate $G_2(ki)$ and $G_2(k\omega)$, where $k\geq2$ and $k\in\mathbb{Z}$. I'll be very appreciated that if someone give the solution.
$f(z)=\frac{(E_2(z)-kE_2(kz))^2}{E_4(z)}$ is a $\Gamma_0(k)$ modular function with rational Fourier coefficients, $f(i)$ is algebraic and radical over $j(i)$.
Concretely looking at finitely many Fourier coefficients you can find the $c_m(z)\in \Bbb{Q}[j(z),j(z)^{-1}]$ such that $$\prod_{\gamma\in \Gamma_0(k)\backslash SL_2(\Bbb{Z})} (X-f(\gamma (z)))=\sum_{m=0}^L c_m(z)X^m $$ The minimal polynomial of $f(i)$ will divide $\sum_{m=0}^L c_m(i)X^m \in \Bbb{Q}[X]$.
It works the same way for $E_2(ke^{2i\pi /3})$, looking at $g(z)=\frac{(E_2(z)-kE_2(kz))^3}{E_6(z)}$.