To solve this problem, let's take it step by step, beginning with calculating each of the vector magnitudes (or norms) and then finding the magnitude of the cross product vector $\overrightarrow{c}$.

Given vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ are:

$\overrightarrow{a} = 3\widehat{i} - 5\widehat{j}$

$\overrightarrow{b} = 6\widehat{i} + 3\widehat{j}$

1. Magnitude of $\overrightarrow{a}$

The magnitude of vector $\overrightarrow{a}$ is calculated using the formula:

$\left| \overrightarrow{a} \right| = \sqrt{(3)^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34}.$

2. Magnitude of $\overrightarrow{b}$

Similarly, the magnitude of vector $\overrightarrow{b}$ is calculated as:

$\left| \overrightarrow{b} \right| = \sqrt{(6)^2 + (3)^2} = \sqrt{36 + 9} = \sqrt{45}.$

3. Calculating $\overrightarrow{c} = \overrightarrow{a} \times \overrightarrow{b}$

The cross product of two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ in 3-dimensional space is given by:

$\overrightarrow{c} = \overrightarrow{a} \times \overrightarrow{b} = \left| \begin{array}{ccc} \widehat{i} & \widehat{j} & \widehat{k} \\ 3 & -5 & 0 \\ 6 & 3 & 0 \end{array} \right|$

For our provided vectors, the $k$ component of both vectors is $0$, thus their cross product will be purely in the $k$ direction. The determinant simplifies to:

$\overrightarrow{c} = (3 \times 3 - (-5) \times 6)\widehat{k} = (9 + 30)\widehat{k} = 39\widehat{k}.$

4. Magnitude of $\overrightarrow{c}$

The magnitude of vector $\overrightarrow{c}$ is:

$\left| \overrightarrow{c} \right| = \sqrt{39^2} = 39.$

Finally, the ratio of the magnitudes of vectors $\overrightarrow{a}$, $\overrightarrow{b}$, and $\overrightarrow{c}$ is thus:

$\sqrt{34}:\sqrt{45}:39.$

Therefore, the correct answer is:

Option B) $\sqrt{34}:\sqrt{45}:39.$