The eccentric angle of the point $\mathrm{P}(-6,2)$ of the ellipse $\frac{x^2}{48}+\frac{y^2}{16}=1$ is

The ellipse is defined by the equation $\frac{x^2}{48}+\frac{y^2}{16}=1$.

By comparing with the standard form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, we find $a=4\sqrt{3}$ and $b=4$.

The parametric coordinates of any point on the ellipse are given by $(a\cos\theta, b\sin\theta)$, where $\theta$ is the eccentric angle.

For the point $\mathrm{P}(-6,2)$, we set:
$4\sqrt{3}\cos\theta = -6$ and $4\sin\theta = 2$.

Solving the second equation gives $\sin\theta = \frac{1}{2}$.

From the first equation, $\cos\theta = \frac{-6}{4\sqrt{3}} = -\frac{\sqrt{3}}{2}$ after rationalization.

Given that $\sin\theta$ is positive and $\cos\theta$ is negative, $\theta$ must lie in the second quadrant.

The reference angle where $\sin\alpha = \frac{1}{2}$ and $\cos\alpha = \frac{\sqrt{3}}{2}$ is $30^{\circ}$.

Thus, $\theta = 180^{\circ} - 30^{\circ} = 150^{\circ}$.

$\boxed{150^{\circ}}$