With the crude approximation that the lift-per-span over the entire wing is a constant, we can estimate (which will be an underestimate in this case) the downwash at the center of the wing as

$LaTeX: \frac{\Gamma}{4\pi}\left(\frac{1}{\frac12 b} + \frac{1}{\frac12 b}\right) = \frac{\Gamma}{\pi b}$

At the center, the effective angle of attack is

$LaTeX: \alpha_e = \alpha + w_i / U_{\infty} = \alpha - \frac{1}{\pi b}\Gamma / U_{\infty}$

Assuming a symmetric, thin airfoil of chord length c, the lift-per-span is then

$LaTeX: 2\pi\,c\,\alpha_e\, \frac12 \rho U_{\infty}^2 = \pi\,c\, \rho U_{\infty}^2\alpha - \frac{c}{b} \rho U_{\infty}\Gamma$

Using Kutta-Joukowski theorem at the center of the wing, we get the linear equation for $LaTeX: \Gamma$ ,

$LaTeX: \rho U_{\infty} \Gamma = \pi\,c\, \rho U_{\infty}^2\alpha - \frac{c}{b} \rho U_{\infty}\Gamma$

the solution of which is

$LaTeX: \Gamma = \frac{\pi\,b\,c}{b+c} U_{\infty}\alpha$

Plug in, the induced downwash at center is then

$LaTeX: \frac{c}{b+c} U_{\infty}\alpha$

The effective angles of attack is

$LaTeX: \frac{b}{b+c} \alpha$

The lift coefficient is

$LaTeX: 2\pi \frac{b}{b+c} \alpha$

This lift is tilted backwards at an angle of $LaTeX: w_i / U_{\infty} = \frac{c}{b+c} \alpha$ . Thus the induced drag coefficient is

$LaTeX: C_l w_i / U_{\infty} = 2\pi\frac{c\,b}{(b+c)^2} \alpha^2$