Let $ABC$ be a triangle in which $\angle ABC = 60^{\circ}$. Let $I$ and $O$ be the incentre and circumcentre of $ABC$, respectively. Let $M$ be the midpoint of the arc $BC$ of the circumcircle of $ABC$, which does not contain the point $A$. Determine $\angle BAC$ given that $MB = OI$.
1. **Trigonometric Approach:**
   - Let \( x = \angle BAC \).
   - Given \( MB = OI \), we infer that \( OM \) is tangent to the circle \((AOI)\). Hence, \( r^2 = OM^2 = MI \cdot MA = MC \cdot MA \).
   - By the Law of Sines, we have:
     \[
     \frac{MC}{\sin{\frac{x}{2}}} = \frac{MA}{\sin{(120^\circ - \frac{x}{2})}} = \frac{AC}{\sin{60^\circ}} = \frac{2\sqrt{3}}{3}AC
     \]
   - Also, \( AC = 2r \cdot \sin{60^\circ} = \sqrt{3}r \).
   - Therefore, we get:
     \[
     4 \cdot \sin{(120^\circ - \frac{x}{2})} \cdot \sin{\frac{x}{2}} = 1
     \]
     \[
     4 \cdot \left( \frac{\sin{\frac{x}{2}}}{2} + \frac{\sqrt{3}}{2} \cos{\frac{x}{2}} \right) \cdot \sin{\frac{x}{2}} = 1
     \]
     \[
     \sqrt{3} \sin{x} + 2 \sin^2{\frac{x}{2}} = 1
     \]
     \[
     \sqrt{3} \sin{x} + \frac{\sin{x} \cdot \sin{\frac{x}{2}}}{\cos{\frac{x}{2}}} = 1
     \]
     \[
     \sqrt{3} + \frac{\sin{\frac{x}{2}}}{\cos{\frac{x}{2}}} = \frac{1}{\sin{x}}
     \]
     \[
     \sqrt{3} + \frac{\sin{x}}{2 \cos^2{\frac{x}{2}}} = \frac{1}{\sin{x}}
     \]
     \[
     \sqrt{3} + \frac{\sin{x}}{\cos{x} + 1} = \frac{1}{\sin{x}}
     \]
     \[
     \sqrt{3} (\cos{x} + 1) \sin{x} + \sin^2{x} = \cos{x} + 1
     \]
     \[
     \sqrt{3} \sin{x} + \sin^2{x} - 1 = \cos{x} - \sqrt{3} \cos{x} \sin{x}
     \]
     \[
     (\sqrt{3} \sin{x} + \sin^2{x} - 1)^2 = (1 - \sqrt{3} \sin{x})^2 (1 - \sin^2{x})
     \]
     \[
     4 \sin^4{x} = \sin^2{x}
     \]
     \[
     \sin{x} = \frac{1}{2}
     \]
     \[
     x = 30^\circ
     \]

2. **Synthetic Approach:**
   - Given \( \angle ABC = 60^\circ \), we have \( \angle CMI = 60^\circ \).
   - Since \( MB = MI = MC \), triangle \( IMC \) is equilateral.
   - The problem statement implies that \( I \) is the circumcenter of \( \triangle OMC \).
   - Let \( \angle BAM = a \). Then \( \angle OIM = 180^\circ - 2a \).
   - Hence, \( \angle OCM = 90^\circ - a \).
   - On the other hand, \( \angle OCM = \angle OCB + \angle BCM = 60^\circ + a \).
   - Therefore, \( a = 15^\circ \).
   - We conclude that \( \angle BAC = 30^\circ \).

The final answer is \(\boxed{30^\circ}\)