TL;DR
it is
$$S_{x_l}(f) = \left\{
\begin{array}{cl}
2N_0, & \vert f \vert < B/2 \\
0, & \text{Otherwise} \\
\end{array}\right.,$$
or
$$S_{x_l}(f) = \left\{
\begin{array}{cl}
N_0, & \vert f \vert < B/2 \\
0, & \text{Otherwise} \\
\end{array}\right.,$$
depends whether you do a normalization, vide my original post
Long question
The answer
Let
$$ x(t) = \text{Re}[x_l(t) e^{j2\pi f_0 t}]$$
be a filtered white Gaussian noise, where $x_l(t) \in \mathbb{C}$ is the complex envelope (also called the lowpass equivalent) of $x(t)$. The Power Spectral Density (PSD) of $x(t)$ is given by
$$S_x(f) = \left\{
\begin{array}{cl}
\frac{N_0}{2}, & \vert f \pm f_0 \vert < B/2 \\
0, & \text{Otherwise} \\
\end{array}\right., \tag{0}$$
where $f_0$ is the carrier frequency and $B$ is the bandwidth of $x(t)$.
Additionally, let $x_i(t)$ and $x_q(t)$ be the phase and quadrature components of $x_l(t)$, i.e., $x_l(t) = x_i(t) + j x_q(t)$. Assuming the stationarity of $x(t)$ leads to the following properties [2, 3]:
$$
R_{x_i}(\tau) = R_{x_q}(\tau) \tag{1}
$$
and
$$
R_{x_i,x_q}(\tau) = -R_{x_q, x_i}(\tau) \tag{2}
$$
Recording that $x(t) = x_i(t) \cos{2\pi f_0 t} - x_q(t) \sin{2\pi f_0 t}$ and using the equations (1) and (2), we have that [2]:
$$
R_{x}(\tau) = R_{x_i}(\tau) \cos{2\pi f_0 \tau} - R_{x_q,x_i}(\tau) \sin{2\pi f_0 \tau} \tag{3}
$$
Since $x_i(t)$ and $x_q(t)$ are independent processes, $R_{x_q,x_i}(\tau) = 0$ and the equation (3) reduces to
$$
\boxed{R_{x}(\tau) = R_{x_i}(\tau) \cos{2\pi f_0 \tau}} \tag{4}
$$
That is the autocorrelation function of the bandpass signal, $x(t)$. But remember that
$$S_x(f) = \left\{
\begin{array}{cl}
\frac{N_0}{2}, & \vert f \pm f_0 \vert < B/2 \\
0, & \text{Otherwise} \\
\end{array}\right.. \tag{5}$$
If you are good at signals and systems, you should already have noticed that (remember that $S_{x_i}(f)$ is the Fourier transform of $R_{x_i}(\tau)$)
$$S_{x_i}(f) = \left\{\begin{array}{cc}
N_0 & \vert f \vert < B/2 \\
0 & \text{Otherwise}.
\end{array}\right. \tag{6}$$
Similarly, recording that $x_l(t) = x_i(t) + j x_q(t)$ and using the equations (1) and (2), we have that (we are normalizing it by $\frac{1}{2}$!!) [1]:
$$
R_{x_l}(\tau) = \frac{1}{2}E[x_l^*(t)x_l(t + \tau)] = R_{x_i}(\tau) + j R_{x_q,x_i}(\tau) \tag{7}
$$
Again, since $x_i(t)$ and $x_q(t)$ are independent, it follows that $R_{x_q,x_i}(\tau) = 0$.
$$
\boxed{R_{x_l}(\tau) = R_{x_i}(\tau)} \tag{7}
$$
Which gives
$$S_{x_l}(f) = \left\{
\begin{array}{cl}
N_0, & \vert f \vert < B/2 \\
0, & \text{Otherwise} \\
\end{array}\right.. \tag{8}$$
Notice that, although $x_l(t)$ is complex, $R_{x_l}(\tau)$ and $S_{x_l}(f)$ are reals [1]!!! More specifically, they are reals and evens: $S_{x_l}(f)$ is square function, and $R_{x_l}(\tau)$ is a sinc.