Answer:

\(\omega=\frac12\sqrt{\frac ga}.\)

Let the tension in the string be \(T\). Since the ring is smooth and the string is light, the tension is the same throughout the string.

Particle \(A\) hangs in equilibrium, so

\(T=mg.\)

For particle \(B\), resolving vertically gives

\(T\cos\theta=\frac12mg.\)

Using \(T=mg\),

\(mg\cos\theta=\frac12mg, \qquad \cos\theta=\frac12.\)

Since \(A\) is the centre of the horizontal circle described by \(B\), \(A\) and \(B\) are at the same vertical level. Hence

\(BR\cos\theta=AR.\)

The total string length is \(12a\), so

\(AR+BR=12a.\)

Using \(\cos\theta=\frac12\), we have \(AR=\frac12BR\). Hence

\(\frac12BR+BR=12a,\)

so

\(BR=8a.\)

For \(B\), resolving horizontally towards the centre of its circle gives

\(T\sin\theta=\frac12m(BR\sin\theta)\omega^2.\)

Cancel \(\sin\theta\):

\(T=\frac12m(BR)\omega^2.\)

Substitute \(T=mg\) and \(BR=8a\):

\(mg=\frac12m(8a)\omega^2.\)

Thus

\(g=4a\omega^2, \qquad \omega^2=\frac{g}{4a}.\)

Therefore

\(\omega=\frac12\sqrt{\frac ga}.\)