First, we prove $a = \frac{(2 \sin 80^{\circ})^2}{\sqrt{3}}$ is sufficient. For any three points $A, B, C$ on the unit circle, let $\angle BAC = \alpha$, $\angle ABC = \beta$, $\angle ACB = \gamma$, $\alpha \le \beta \le \gamma$.

If $\beta \le 60^{\circ}$, since $AB \le 2 < a$ (here, $a > \frac{(2 \sin 75^{\circ})^2}{\sqrt{3}} = \frac{(\sqrt{6} + \sqrt{2})^2}{4\sqrt{3}} = \frac{8 + 4\sqrt{3}}{4\sqrt{3}} > 2$), one can draw a segment $PQ$ of length $a$ that contains $A$ and $B$, then find $R$ such that $\triangle PQR$ is equilateral, $C$, $R$ on the same side of $PQ$. It follows from $\alpha \le \beta \le 60^{\circ}$ that $C$ is inside $\triangle PQR$ or on the boundary.

If $\beta > 60^{\circ}$, we first prove the following:
At least one of $\sin \beta \sin \gamma$ and $\sin \beta \sin(\alpha + 60^{\circ})$ is less than or equal to $\sin^2 80^{\circ}$. (*)

In fact, as $2(\beta + \alpha + 60^{\circ}) + (\beta + \gamma) = 480^{\circ} - (\gamma - \beta) \le 480^{\circ}$, one of $\beta + \alpha + 60^{\circ}$ and $\beta + \gamma$ does not exceed $160^{\circ}$. From the simple property of sine function
$$\begin{aligned} \sin x \sin y &= \frac{1}{2}(\cos(x - y) - \cos(x + y)) \\ &\le \frac{1}{2}(1 - \cos(x + y)) \\ &= \sin^2 \frac{x+y}{2}, \end{aligned}$$
(*) follows.

i) If $\sin \beta \sin \gamma \le \sin^2 80^{\circ}$, let $AD \perp BC$ with foot $D$; take $P, Q$ on the line $BC$ such that $PD = DQ = \frac{a}{2}$; take $R$ on the ray $DA$ such that
$$DR = \frac{\sqrt{3}a}{2} = 2 \sin^2 80^{\circ}.$$
Notice that $AD = 2 \sin \beta \sin \gamma \le DR$, and thus $A$ lies inside $\triangle PQR$ or on the boundary; from $60^{\circ} \le \beta, \gamma \le 120^{\circ}$, it is easy to see that $B, C$ lie on the segment $PQ$.

ii) If $\sin \beta \sin(\alpha + 60^{\circ}) \le \sin^2 80^{\circ}$, take $P = A$, extend $AB$ to $Q$, $AQ = a$, and find $R$ such that $\triangle PQR$ is equilateral and $C, R$ lie on the same side of $PQ$. Let $M$ be the intersection of the ray $AC$ and the line $QR$. Notice that $AB \le 2 < a$, hence $B$ is on the boundary of $\triangle PQR$; since
$$AM = a \cdot \frac{\sin 60^{\circ}}{\sin(\alpha + 60^{\circ})} = \frac{2 \sin^2 80^{\circ}}{\sin(\alpha + 60^{\circ})} > 2 \sin \beta = AC,$$
$C$ is inside $\triangle PQR$ or on the boundary.

In conclusion, $a = \frac{(2 \sin 80^{\circ})^2}{\sqrt{3}}$ satisfies the problem condition.

Next, we prove $a \ge \frac{(2 \sin 80^{\circ})^2}{\sqrt{3}} = \frac{2 \sin^2 80^{\circ}}{\sin 60^{\circ}}$ is necessary. To this end, choose $A, B, C$ on the unit circle with $\angle BAC = 20^{\circ}$, $\angle ABC = \angle ACB = 80^{\circ}$.

Suppose the equilateral triangle $PQR$ with side length $a$ covers $A, B, C$ (inside or on the boundary). By a translation if necessary, we can make one of $A, B, C$, say $A$ on the boundary of $\triangle PQR$, and then by a rotation about $A$ we can make $B$ or $C$ on the boundary. If neither of them is a vertex of $\triangle PQR$, assume they all lie on $PQ$ or $PR$. Take $P$ as the homothetic centre and rescale $\triangle PQR$ smaller such that all of $A, B, C$ fall on the boundary or become a vertex of $\triangle PQR$. Since $\angle ABC, \angle ACB > 60^{\circ}$, $B, C$ cannot be vertices; in addition, there is symmetry between $B$ and $C$. Thus, it suffices to consider four situations as follows.

(1) $A$ is a vertex, say $A = P$. Let $PH \perp QR$ with foot $H$. Then the angle between $PH$ and one of $PB, PC$ is less than or equal to $10^{\circ}$. It follows that $PH \ge 2 \sin^2 80^{\circ}$, and
$$a = PQ \geq \frac{2 \sin^2 80^{\circ}}{\sin 60^{\circ}}.$$

(2) $A$ and $B$ lie on the same side, say $PQ$, with $PA \le PB$. Let $PH \perp QR$ at $H$. Then the angle between $AC$ and $PH$ is $10^{\circ}$. Similar to (1), we have
$$a \geq \frac{2 \sin^2 80^{\circ}}{\sin 60^{\circ}}.$$

(3) $B$ and $C$ lie on the same side, say $QR$. Let $PH \perp QR$ at $H$. Then the lines $AC$ and $PH$ cross at $10^{\circ}$. Similar to (1), we have
$$a \geq \frac{2 \sin^2 80^{\circ}}{\sin 60^{\circ}}.$$

(4) $A, B, C$ are all on different sides. Let $A \in PQ$, $B \in QR$, $C \in RP$, and $\vartheta = \angle RBC \le \angle RCB$, $\vartheta \le 60^{\circ}$.

i) If $\vartheta \le 20^{\circ}$, then the angle between $AB$ and the altitude on $QR$ does not exceed $10^{\circ}$. Similar to (1), we get
$$a \geq \frac{2 \sin^2 80^{\circ}}{\sin 60^{\circ}}.$$

ii) If $40^{\circ} \le \vartheta \le 60^{\circ}$, then the angle between $AC$ and the altitude on $PQ$ does not exceed $10^{\circ}$. Similar to (1), $a \ge \frac{2 \sin^2 80^{\circ}}{\sin 60^{\circ}}$.

iii) If $20^{\circ} \le \vartheta \le 40^{\circ}$, then
$$\begin{aligned} a &= QR = QB + BR \\ &= \frac{2 \sin 80^{\circ} \sin(20^{\circ} + \vartheta)}{\sin 60^{\circ}} + \frac{2 \sin 20^{\circ} \sin(60^{\circ} + \vartheta)}{\sin 60^{\circ}} \\ &\ge \frac{2 \sin 80^{\circ} \sin 40^{\circ} + 2 \sin 20^{\circ} \sin 80^{\circ}}{\sin 60^{\circ}} \\ &= \frac{2 \sin^2 80^{\circ}}{\sin 60^{\circ}}. \end{aligned}$$

In summary, the minimum is $a = \frac{2 \sin^2 80^{\circ}}{\sin 60^{\circ}}$. $\square$