The correct answer is $\boxed{{A}{e^{ – B{f^2}}}}$.
The Fourier Transform of a function $x(t)$ is defined as:
$$X(f) = \int_{-\infty}^{\infty} x(t) e^{-j2\pi ft} dt$$
In this case, $x(t) = {e^{ – 3{t^2}}}$. Substituting this into the Fourier Transform formula, we get:
$$X(f) = \int_{-\infty}^{\infty} {e^{ – 3{t^2}}} e^{-j2\pi ft} dt$$
We can use the following identity to evaluate this integral:
$$\int_{-\infty}^{\infty} {e^{ – a{t^2}}} e^{-j2\pi ft} dt = \sqrt{\frac{\pi}{a}} e^{-B{f^2}}$$
where $B = \frac{2\pi}{a}$.
Substituting this into the Fourier Transform formula, we get:
$$X(f) = \sqrt{\frac{\pi}{3}} e^{-B{f^2}}$$
Therefore, the Fourier Transform of the signal $x(t) = {e^{ – 3{t^2}}}$ is of the form ${A}{e^{ – B{f^2}}}$, where $A = \sqrt{\frac{\pi}{3}}$.
Option A is incorrect because it does not include the factor of $\sqrt{\frac{\pi}{3}}$.
Option B is incorrect because it does not include the factor of $e^{-Bt^2}$.
Option C is incorrect because it is not a Fourier Transform.
Option D is incorrect because it does not include the factor of $e^{-Bt^2}$.