Step-by-step Solution:

1. Understand the Complex Number and its Powers

   We are given the complex number $z = \cos \theta + i \sin \theta$. By Euler's formula, this can be written as $z = e^{i\theta}$.

   We need to find the imaginary part of $z^{2m-1}$. Using De Moivre's Theorem, which states that $(\cos \theta + i \sin \theta)^n = \cos(n\theta) + i \sin(n\theta)$, we have: $z^{2m-1} = (\cos \theta + i \sin \theta)^{2m-1} = \cos((2m-1)\theta) + i \sin((2m-1)\theta)$

2. Identify the Imaginary Part

   The imaginary part of $z^{2m-1}$, denoted by ${\mathop{\rm Im} olimits} ({z^{2m - 1}})$, is the coefficient of $i$. ${\mathop{\rm Im} olimits} ({z^{2m - 1}}) = \sin((2m-1)\theta)$

3. Formulate the Sum

   The problem asks for the value of the sum $S = \sum\limits_{m = 1}^{15} {{\mathop{\rm Im} olimits} } ({z^{2m - 1}})$. Substituting the expression from Step 2, we get: $S = \sum_{m=1}^{15} \sin((2m-1)\theta)$

   Let's expand the sum to see the terms: $S = \sin(\theta) + \sin(3\theta) + \sin(5\theta) + \dots + \sin(29\theta)$

4. Evaluate the Trigonometric Series

   This is a sum of sine terms where the angles are in an arithmetic progression: $\theta, 3\theta, 5\theta, \dots$. The first term is $A = \theta$, the common difference is $D = 2\theta$, and the number of terms is $n = 15$.

   The formula for the sum of a sine series is: $\sum_{k=1}^{n} \sin(A + (k-1)D) = \frac{\sin(nD/2)}{\sin(D/2)} \sin\left(A + \frac{(n-1)D}{2}\right)$

   Applying this formula with our values:

   • $n = 15$
   • $D = 2\theta$
   • $A = \theta$

   We get:

   • $nD/2 = 15(2\theta)/2 = 15\theta$
   • $D/2 = 2\theta/2 = \theta$
   • $A + \frac{(n-1)D}{2} = \theta + \frac{(15-1)2\theta}{2} = \theta + 14\theta = 15\theta$

   Substituting these into the formula: $S = \frac{\sin(15\theta)}{\sin(\theta)} \sin(15\theta) = \frac{\sin^2(15\theta)}{\sin(\theta)}$

5. Substitute the Given Value of θ

   We are given that $\theta = 2^\circ$. Substituting this into the expression for S: $S = \frac{\sin^2(15 \times 2^\circ)}{\sin(2^\circ)} = \frac{\sin^2(30^\circ)}{\sin(2^\circ)}$

6. Calculate the Final Value

   We know that $\sin(30^\circ) = \frac{1}{2}$. Therefore, $\sin^2(30^\circ) = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$.

   Substituting this value back into the expression for S: $S = \frac{1/4}{\sin(2^\circ)} = \frac{1}{4\sin(2^\circ)}$

7. Conclusion

   The value of the sum is $\frac{1}{4\sin(2^\circ)}$. Comparing this with the given options: A: ${1 \over {\sin \,{2^ \circ }}}$ B: ${1 \over {3\sin \,{2^ \circ }}}$ C: ${1 \over {2\sin \,{2^ \circ }}}$ D: ${1 \over {4\sin \,{2^ \circ }}}$

   The calculated value matches option D.