1. Set up the standard ellipse relations

For an ellipse with semi-major axis $a$, semi-minor axis $b$, and eccentricity $e$:

$e=\frac{c}{a}, \qquad c^2=a^2-b^2$

The length of the latus rectum is

$\frac{2b^2}{a}$

Given:

$\frac{2b^2}{a}=4$

So,

$b^2=2a \quad \text{(1)}$

2. Use the distance between focus and nearest vertex

The focus is at distance $c=ae$ from the center, and the nearest vertex on the major axis is at distance $a$ from the center.

So the distance between a focus and its nearest vertex is:

$a-c=a-ae=a(1-e)$

Given this distance is $\frac{3}{2}$:

$a(1-e)=\frac{3}{2} \quad \text{(2)}$

3. Relate $b^2$ and $e$

Since

$b^2=a^2(1-e^2)$

and from (1), $b^2=2a$, we get

$a^2(1-e^2)=2a$

Assuming $a\neq 0$, divide by $a$:

$a(1-e^2)=2 \quad \text{(3)}$

4. Substitute for $a$ using (2)

From (2):

$a=\frac{3}{2(1-e)}$

Put this into (3):

$\frac{3}{2(1-e)}(1-e^2)=2$

Now,

$1-e^2=(1-e)(1+e)$

So,

$\frac{3}{2}(1+e)=2$

$3(1+e)=4$

$1+e=\frac{4}{3}$

$e=\frac{1}{3}$

5. Check with options

The eccentricity is

$\boxed{\frac{1}{3}}$

So the correct option is B.