We will use the fact that
$\begin{align}\int_{\infty}^{\infty} e^{- (x - a)^2} dx = \sqrt{\pi} \label{eqn:1}\end{align}$
where $a$ is any complex number.
We assume that $f(k)$ is a Gaussian and that $\Delta k$ is its standard deviation. Then
$\begin{align*}f(k) = \frac{1}{\sqrt{2 \pi \Delta k ^2}} e^{- (k - k_0)^2/2\Delta k^2}\end{align*}$
where the prefactor of $1/\sqrt{2 \pi \Delta k ^2}$ is for normalization. Then,
$\begin{align*}\psi(x,t) &= \int_{-\infty}^{\infty} e^{i (k x - \omega t)} \frac{1}{\sqrt{2 \pi \Delta k ^2}} e^{- (k - k_...$
where equation (1) was used to evaluate the integral.
This implies that
$\begin{align*}\boxed{\Delta k = \frac{1}{\Delta x_0}}\end{align*}$
Therefore, answer (B) is correct.

Solution to 1992 Problem 27