Answer
$I_1 = \frac{\pi D}{R} I_2$
Work Step by Step
$B = B_{loop} + B_{wire} = 0$
$B_{loop} = -\mu_0 \frac{I_2}{2R}$ (Field inside a coil of wire)
$B_{wire} = \mu_0 \frac{I_1}{2\pi D}$ (Ampere's law)
So $\frac{\mu_0}{2} \left(\frac{I_1}{\pi D} - \frac{I_2}{R}\right) = 0$ and $I_1 = \frac{\pi D}{R} I_2$
