Let $M$ be the intersection of the diagonals $AC$ and $BD$ of cyclic quadrilateral $ABCD$. If $|AB|=5$, $|CD|=3$, and $m(\widehat{AMB}) = 60^\circ$, what is the circumradius of the quadrilateral?

$ 
\textbf{(A)}\ 5\sqrt 3
\qquad\textbf{(B)}\ \dfrac {7\sqrt 3}{3}
\qquad\textbf{(C)}\ 6
\qquad\textbf{(D)}\ 4
\qquad\textbf{(E)}\ \sqrt{34}
$
1. **Identify the given information and the goal:**
   - We are given a cyclic quadrilateral \(ABCD\) with diagonals \(AC\) and \(BD\) intersecting at \(M\).
   - The lengths of sides \(AB\) and \(CD\) are given as \(5\) and \(3\) respectively.
   - The angle \(\angle AMB\) is given as \(60^\circ\).
   - We need to find the circumradius \(R\) of the quadrilateral.

2. **Use the properties of cyclic quadrilaterals:**
   - In a cyclic quadrilateral, the opposite angles sum up to \(180^\circ\).
   - The angle \(\angle AMB\) is formed by the intersection of diagonals, and it is given as \(60^\circ\).

3. **Apply the Law of Cosines in \(\triangle AMB\):**
   - Since \(\angle AMB = 60^\circ\), we can use the Law of Cosines to find the length of \(BD\).
   - Let \(BD = x\).
   - Using the Law of Cosines in \(\triangle AMB\):
     \[
     x^2 = AB^2 + AM^2 - 2 \cdot AB \cdot AM \cdot \cos(60^\circ)
     \]
     Since \(\cos(60^\circ) = \frac{1}{2}\), we have:
     \[
     x^2 = 5^2 + AM^2 - 2 \cdot 5 \cdot AM \cdot \frac{1}{2}
     \]
     Simplifying, we get:
     \[
     x^2 = 25 + AM^2 - 5 \cdot AM
     \]

4. **Use the properties of cyclic quadrilaterals and the given angle:**
   - Since \(M\) is the intersection of the diagonals, we can use the fact that the diagonals of a cyclic quadrilateral bisect each other.
   - Therefore, \(AM = CM\) and \(BM = DM\).

5. **Calculate the circumradius \(R\):**
   - The circumradius \(R\) of a cyclic quadrilateral can be found using the formula:
     \[
     R = \frac{\sqrt{(AB \cdot CD + AD \cdot BC)(AB \cdot CD + AD \cdot BC - AC \cdot BD)}}{4 \cdot \text{Area of } ABCD}
     \]
   - However, we need to find the lengths of \(AC\) and \(BD\) to use this formula.

6. **Simplify the problem using the given angle:**
   - Since \(\angle AMB = 60^\circ\), we can use the fact that the diagonals of a cyclic quadrilateral intersect at an angle that is supplementary to the angle formed by the opposite sides.
   - Therefore, \(\angle ACD = 120^\circ\).

7. **Use the Law of Cosines in \(\triangle ACD\):**
   - Let \(AC = y\).
   - Using the Law of Cosines in \(\triangle ACD\):
     \[
     y^2 = AD^2 + CD^2 - 2 \cdot AD \cdot CD \cdot \cos(120^\circ)
     \]
     Since \(\cos(120^\circ) = -\frac{1}{2}\), we have:
     \[
     y^2 = AD^2 + 3^2 + 2 \cdot AD \cdot 3 \cdot \frac{1}{2}
     \]
     Simplifying, we get:
     \[
     y^2 = AD^2 + 9 + 3 \cdot AD
     \]

8. **Find the circumradius \(R\):**
   - Using the simplified formula for the circumradius of a cyclic quadrilateral:
     \[
     R = \frac{AB \cdot CD}{2 \cdot \sin(\angle AMB)}
     \]
     Substituting the given values:
     \[
     R = \frac{5 \cdot 3}{2 \cdot \sin(60^\circ)}
     \]
     Since \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\), we have:
     \[
     R = \frac{15}{2 \cdot \frac{\sqrt{3}}{2}} = \frac{15}{\sqrt{3}} = 5\sqrt{3}
     \]

The final answer is \(\boxed{5\sqrt{3}}\)