Answer
$B = - \frac{\mu_0 I}{4R}$
Work Step by Step
Biot-Savart law : $\vec{dB} = \frac{\mu_0}{4\pi} \frac{I \vec{dl} \times \hat{r}}{r^2}$
With $\vec{l} = \pi R$, $\hat{r}$ perpendicular to $\vec{l}$ and $r = R$, we find :
$B = \frac{-\mu_0}{4\pi} \frac{I \pi R}{R^2} = -\frac{\mu_0 I}{4R}$
The minus sign comes from the cross product and depends on a you define the vectors.
