1. Write the standard form of the hyperbola

Since the centre is at the origin and the transverse axis is along the $x$-axis, the hyperbola is of the form

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1.$

2. Use the transverse axis length

For this hyperbola, the length of the transverse axis is $2a$. Given:

$2a=4 \implies a=2 \implies a^2=4.$

So the equation becomes

$\frac{x^2}{4}-\frac{y^2}{b^2}=1.$

3. Use the fact that the hyperbola passes through $(4,2)$

Substitute $(x,y)=(4,2)$:

$\frac{4^2}{4}-\frac{2^2}{b^2}=1$ $\frac{16}{4}-\frac{4}{b^2}=1$ $4-\frac{4}{b^2}=1$ $\frac{4}{b^2}=3$ $b^2=\frac{4}{3}.$

4. Find the eccentricity

For the hyperbola

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,$

we have

$c^2=a^2+b^2$

and eccentricity

$e=\frac{c}{a}.$

Now,

$c^2=4+\frac{4}{3}=\frac{16}{3}$ $c=\frac{4}{\sqrt{3}}.$

Hence,

$e=\frac{c}{a}=\frac{\frac{4}{\sqrt{3}}}{2}=\frac{2}{\sqrt{3}}.$

5. Match with the options

$\boxed{\frac{2}{\sqrt{3}}}$

So the correct option is D.