To find the radius of the circle given the points \(A\), \(B\), and \(C\) on the circle with \(AB = 3 \text{ cm}\) and \(BC = 4 \text{ cm}\), we must consider the properties of a circle and the triangle formed by these points.

Assuming \(A\), \(B\), and \(C\) are points on the circle, the triangle \(ABC\) forms a right-angled triangle inscribed in the circle, with the hypotenuse as the diameter of the circle. This is based on Thales' theorem.

Since \( \angle ACB = 90^\circ \), we can treat \(AC\) as the hypotenuse of the triangle \(ABC\). Using the Pythagorean theorem:

\(AC^2 = AB^2 + BC^2\)

Substitute the given values:

\(AC^2 = 3^2 + 4^2 = 9 + 16 = 25\)

Therefore, \(AC = \sqrt{25} = 5 \text{ cm}\).

Since \(AC\) is the diameter of the circle (from \(A\) to \(C\)), the radius \(r\) is half of the diameter.

\(r = \frac{AC}{2} = \frac{5}{2} \text{ cm}\)

Thus, the measure of the radius of the circle is \(\frac{5}{2} \text{ cm}\).

The correct answer is \(\frac{5}{2} \text{ cm}\).