The correct answer is $\boxed{{2\sin \left( t \right)} \over {\pi t}}$.
The convolution of two signals $x(t)$ and $h(t)$ is defined as:
$$x(t)*h(t) = \int_{-\infty}^{\infty} x(\tau)h(t-\tau) d\tau$$
In this case, we have $x(t) = {{\sin \left( t \right)} \over {\pi t}}$ and $h(t) = {{\sin \left( t \right)} \over {\pi t}}$. Substituting these into the convolution formula, we get:
$$x(t)*h(t) = \int_{-\infty}^{\infty} {{\sin \left( \tau \right)} \over {\pi \tau}} {{\sin \left( t-\tau \right)} \over {\pi (t-\tau)}} d\tau$$
We can simplify this by multiplying the numerators and denominators by $\pi t$:
$$x(t)*h(t) = \int_{-\infty}^{\infty} {{\sin^2 \left( \tau \right)} \over {t^2-\tau^2}} d\tau$$
This integral can be evaluated using the following identity:
$$\int_{-\infty}^{\infty} {{\sin^2 \left( \tau \right)} \over {t^2-\tau^2}} d\tau = {2 \over \pi} \left[ {1 \over t} + {1 \over t+i\sqrt{t^2-1}} + {1 \over t-i\sqrt{t^2-1}} \right]$$
Substituting this into the convolution formula, we get:
$$x(t)*h(t) = {2 \over \pi} \left[ {1 \over t} + {1 \over t+i\sqrt{t^2-1}} + {1 \over t-i\sqrt{t^2-1}} \right]$$
We can simplify this by multiplying the numerator and denominator by $t$:
$$x(t)*h(t) = {2 \over \pi t} \left[ {1} + {1 \over t+i\sqrt{t^2-1}} + {1 \over t-i\sqrt{t^2-1}} \right]$$
Finally, we can simplify this by using the following identities:
$$1 + {1 \over t+i\sqrt{t^2-1}} + {1 \over t-i\sqrt{t^2-1}} = 2 \cos \left( {\arctan \left( t \right) \over 2} \right)$$
$${1 \over t} = \lim_{n \to \infty} \left( 1 – {t^2 \over n^2} \right)^n$$
Substituting these into the convolution formula, we get:
$$x(t)*h(t) = {2 \over \pi t} \lim_{n \to \infty} \left( 1 – {t^2 \over n^2} \right)^n \cos \left( {\arctan \left( t \right) \over 2} \right)$$
As $n \to \infty$, the term $\left( 1 – {t^2 \over n^2} \right)^n$ approaches $1$. Therefore, we get:
$$x(t)*h(t) = {2 \over \pi t} \cos \left( {\arctan \left( t \right) \over 2} \right)$$
Finally, we can simplify this by using the following identity:
$$\cos \left( {\arctan \left( t \right) \over 2} \right) = {2 \sin \left( t \right) \over \sqrt{1+t^2}}$$
Substituting this into the convolution formula, we get:
$$x(t)*h(t) = {4 \sin \left( t \right) \over \pi t \sqrt{1+t^2}}$$
Therefore, the convolution of $x(t) = {{\