To find the angle made by the line \(L\) with the positive direction of the y-axis, we begin by analyzing the equations given.

The line \(L\) is given by the equation: \(x + \sqrt{3} y + 3\sqrt{3} = 0\).

Let's first rewrite the line equation in the slope-intercept form, \(y = mx + c\). Rearranging, we have:

\(\sqrt{3}y = -x - 3\sqrt{3}\)

Divide through by \(\sqrt{3}\) to solve for \(y\):

\(y = -\frac{1}{\sqrt{3}}x - 3\)

This gives us the slope \(m = -\frac{1}{\sqrt{3}}\).

The slope of a line makes an angle \(\theta\) with the positive x-axis such that \(\tan \theta = m\). For our line:

\(\tan \theta = -\frac{1}{\sqrt{3}}\)

We know that \(\tan 30^\circ = \frac{1}{\sqrt{3}}\), hence:

\(\theta = 180^\circ - 30^\circ = 150^\circ\)

Now, to find the angle \(\beta\) made with the positive y-axis, note that:

Angle with y-axis = \(90^\circ\) - any angle with x-axis in positive x-plane if the line moves counter-clockwise.

Thus, the angle made by line \(L\) with the positive direction of y-axis is:

\(\beta = 180^\circ - (90^\circ + 30^\circ) = 60^\circ\)

Therefore, the correct answer is \(60^\circ\).