The velocity of the boat as observed from the shore ($\vec{v}_{bs}$) is the vector sum of the boat's velocity relative to the water ($\vec{v}_{bw}$) and the velocity of the water relative to the shore ($\vec{v}_{ws}$), which is the current.

$$ \vec{v}_{bs} = \vec{v}_{bw} + \vec{v}_{ws} $$

Let's establish a coordinate system where the positive x-axis points East and the positive y-axis points North.

The velocity of the river current is purely eastward:

$$ \vec{v}_{ws} = (3.0 \text{ m/s}) \hat{i} $$

The velocity of the boat relative to the water is directed 30° east of north. Let $v_{bw} = 2.0$ m/s and the angle from the North (y-axis) towards the East (x-axis) be $\theta = 30^\circ$. We decompose this vector into its components:

$$ v_{bw,x} = v_{bw} \sin\theta $$ $$ v_{bw,y} = v_{bw} \cos\theta $$ $$ \vec{v}_{bw} = (v_{bw} \sin\theta) \hat{i} + (v_{bw} \cos\theta) \hat{j} $$

The resultant velocity of the boat relative to the shore, $\vec{v}_{bs}$, is found by adding the components:

$$ \vec{v}_{bs} = (v_{bw} \sin\theta + v_{ws}) \hat{i} + (v_{bw} \cos\theta) \hat{j} $$

The magnitude of the resultant velocity is:

$$ |\vec{v}_{bs}| = \sqrt{(v_{bw} \sin\theta + v_{ws})^2 + (v_{bw} \cos\theta)^2} $$

The direction, specified by the angle $\phi$ north of east, is:

$$ \phi = \arctan\left(\frac{v_{bs,y}}{v_{bs,x}}\right) = \arctan\left(\frac{v_{bw} \cos\theta}{v_{bw} \sin\theta + v_{ws}}\right) $$

Now, we substitute the given values: $v_{ws} = 3.0$ m/s, $v_{bw} = 2.0$ m/s, and $\theta = 30^\circ$.

First, calculate the components of the resultant velocity:

$$ v_{bs,x} = (2.0 \text{ m/s}) \sin(30^\circ) + 3.0 \text{ m/s} = (2.0)(0.5) + 3.0 = 4.0 \text{ m/s} $$ $$ v_{bs,y} = (2.0 \text{ m/s}) \cos(30^\circ) = (2.0)\left(\frac{\sqrt{3}}{2}\right) = \sqrt{3} \approx 1.732 \text{ m/s} $$

Next, calculate the magnitude:

$$ |\vec{v}_{bs}| = \sqrt{(4.0 \text{ m/s})^2 + (\sqrt{3} \text{ m/s})^2} = \sqrt{16 + 3} \text{ m/s} = \sqrt{19} \text{ m/s} \approx 4.4 \text{ m/s} $$

Finally, calculate the direction:

$$ \phi = \arctan\left(\frac{\sqrt{3} \text{ m/s}}{4.0 \text{ m/s}}\right) \approx 23.4^\circ $$

The direction is $23^\circ$ North of East.