The correct answer is $\boxed{\text{B) }e^{-f}u(f)}$.
The Fourier transform of a function $f(t)$ is defined as
$$F(f) = \int_{-\infty}^{\infty} f(t) e^{-j2\pi ft} dt$$
The inverse Fourier transform is defined as
$$f(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} F(f) e^{j2\pi ft} df$$
The Fourier transform of a unit step function $u(t)$ is
$$F(f) = \begin{cases} 1 & \text{if } f = 0 \\ 0 & \text{if } f \neq 0 \end{cases}$$
The Fourier transform of an exponential function $e^{-at}$ is
$$F(f) = \frac{1}{a + j2\pi f}$$
Therefore, the Fourier transform of $e^{-t}u(t)$ is
$$F(f) = \frac{1}{1 + j2\pi f}$$
The inverse Fourier transform of $e^{-f}u(f)$ is
$$f(t) = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{-f}u(f) e^{j2\pi ft} df = e^{-t}u(t)$$
Therefore, the Fourier transform of ${{1 \over {1 + j2\pi f}}}$ is $e^{-f}u(f)$.