To find the value of \(QR\), we will use the properties of the given figure.

1. The triangle \(RPS\) is an equilateral triangle, which means all sides are equal and all angles are \(60^\circ\).
2. Thus, the side \(RS = 3\sqrt{3} \text{ cm}\) implies that \(RP = PS = 3\sqrt{3} \text{ cm}\) as well.
3. In the triangle \(PQS\), we see that angle \(\angle PQS = 90^\circ\) and \(\angle PSQ = 60^\circ\), making it a 30-60-90 right triangle.
4. In a 30-60-90 triangle, the side ratios opposite these angles are \(1:\sqrt{3}:2\).
5. Let \(PQ = x\). Then \(QS = \sqrt{3}x\), and the hypotenuse, \(PS = 2x\).
6. Since \(PS = 3\sqrt{3} \text{ cm}\), equating gives: \(2x = 3\sqrt{3}\)
7. Solving for \(x\): \(x = \frac{3\sqrt{3}}{2}\)
8. Thus, \(PQ = \frac{3\sqrt{3}}{2} \text{ cm}\).
9. Since \(\angle PQR = 60^\circ\), we can establish another triangle \(PQR\), where: \(QR^2 = PQ^2 + PR^2 - 2 \cdot PQ \cdot PR \cdot \cos(60^\circ)\)
10. We already know \(PR = 3\sqrt{3} \text{ cm}\) and \(\cos(60^\circ) = \frac{1}{2}\)
11. Plug these into the equation: 

1. Calculate \(QR^2\): 

1. Simple calculations give: 

1. Thus, \(QR = \sqrt{6} = 6 \text{ cm}\).

Hence, the correct answer is \(\text{6 cm}\).