The correct answer is $\boxed{\text{A. Cannot be determined}}$.
The Laplace transform of a function $f(t)$ is defined as
$$L[f(t)] = \int_0^\infty f(t) e^{-st} dt$$
The Laplace transform of a constant function is $1$. Therefore, if $sL[f(t)] = \frac{\omega}{\left(s^2 + \omega^2\right)}$, then
$$\int_0^\infty f(t) e^{-st} dt = \frac{\omega}{\left(s^2 + \omega^2\right)}$$
This equation can be solved for $f(t)$ in terms of the inverse Laplace transform, which is defined as
$$f(t) = \frac{1}{2\pi i} \int_{c-i\infty}^{c+i\infty} L[f(t)] e^{st} ds$$
where $c$ is a real number greater than any of the poles of $L[f(t)]$.
In this case, the poles of $L[f(t)]$ are at $s = \pm i\omega$. Therefore, the inverse Laplace transform is
$$f(t) = \frac{1}{2\pi i} \int_{-i\omega}^{i\omega} \frac{\omega}{\left(s^2 + \omega^2\right)} e^{st} ds$$
This integral cannot be evaluated in closed form. Therefore, the value of $f(t)$ cannot be determined.