1. Interpret the given circle

The circle $x^2+(y-1)^2=1$ has centre $(0,1)$ and radius $1$.

Since it touches the $x$-axis at $(0,0)$, any circle touching it externally and also touching the $x$-axis will have centre $(\alpha,\beta)$ with radius $\beta$ (because distance from the centre to the $x$-axis equals the radius, and $\beta>0$).

2. Condition for external touching

Let the required circle have centre $(\alpha,\beta)$ and radius $\beta$.

For external tangency with the given circle, $\text{distance between centres} = \text{sum of radii}.$

So, $\sqrt{(\alpha-0)^2+(\beta-1)^2}=\beta+1.$

Squaring, $\alpha^2+(\beta-1)^2=(\beta+1)^2.$

Expand both sides: $\alpha^2+\beta^2-2\beta+1=\beta^2+2\beta+1.$

Hence, $\alpha^2=4\beta.$

So the locus $L$ is $\beta=\frac{\alpha^2}{4}, \quad \beta>0,$ which is the parabola $y=\frac{x^2}{4}.$

3. Find the region bounded by $L$ and $y=4$

We need the area enclosed between $y=\frac{x^2}{4} \quad \text{and} \quad y=4.$

First find the points of intersection: $4=\frac{x^2}{4} \implies x^2=16 \implies x=\pm 4.$

So the required area is $A=\int_{-4}^{4}\left(4-\frac{x^2}{4}\right)dx.$

4. Evaluate the integral

$A=\int_{-4}^{4}4\,dx-\frac14\int_{-4}^{4}x^2\,dx.$

Now, $\int_{-4}^{4}4\,dx=4(8)=32.$

Also, $\int_{-4}^{4}x^2\,dx=2\int_0^4 x^2\,dx=2\left[\frac{x^3}{3}\right]_0^4=2\cdot \frac{64}{3}=\frac{128}{3}.$

Therefore, $A=32-\frac14\cdot \frac{128}{3}=32-\frac{32}{3}=\frac{96-32}{3}=\frac{64}{3}.$

5. Match with the options

$\boxed{\frac{64}{3}}$

So the correct option is C.