The correct answer is $\boxed{{1 \over 2}}$.
The Fourier transform of a signal $h(t)$ is defined as
$$H\left( {j\omega } \right) = \int_{-\infty}^{\infty} h(t) e^{-j\omega t} dt$$
The inverse Fourier transform is defined as
$$h(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} H\left( {j\omega } \right) e^{j\omega t} d\omega$$
In this case, we are given that
$$H\left( {j\omega } \right) = {{\left( {2\cos \omega } \right)\left( {\sin \omega } \right)} \over \omega }$$
The inverse Fourier transform of this function can be found using the following formula:
$$h(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} {{\left( {2\cos \omega } \right)\left( {\sin \omega } \right)} \over \omega } e^{j\omega t} d\omega$$
We can evaluate this integral using the following steps:
- We can use the substitution $u = \omega t$ to write the integral as
$$h(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} {{\left( {2\cos u } \right)\left( {\sin u } \right)} \over u } e^{-u} du$$
- We can then use the following formula to evaluate the integral:
$$\int_{-\infty}^{\infty} {{\left( {2\cos u } \right)\left( {\sin u } \right)} \over u } e^{-u} du = \pi$$
- Therefore, we have
$$h(t) = \frac{1}{2\pi} \pi = \frac{1}{2}$$
Therefore, the value of $h(0)$ is $\boxed{{1 \over 2}}$.