What happens to the standard deviation of a white Gaussian noise after low pass filter?
I assume you are working with discrete-time, since continuous-time white noise has infinite power ($\sigma^2$).

First, remember that the power of a stationary process is always equal to the autocorrelation at 0 ($P_x = R_x[0]$); and the variance is the autocovariance at 0 ($\sigma^2_x = \rm{Cov}_x[0]$). These 2 expressions are equal for processes with 0 mean, which is the case for white noise. So, if you already calculated the autocorrelation, you have the power.

But you don't need to calculate the autocorrelation of the filtered signal. You can use Parseval's property and calculate the variance in frequency domain:

$\sigma_y^2 = \frac{1}{2\pi}\int_{2\pi} |H(e^{j\theta})|^2 G_x(e^{j\theta}) d\theta = \frac{1}{2\pi}\int_{-\theta_0}^{+\theta_0} 1 \times \sigma^2 d\theta$ since the ideal lowpass filter is 1 between $\pm\theta_0$, and the spectral density of the input signal is $\sigma^2$ (white noise).

The result is very easy to calculate, and gives $\sigma_y^2 = (\theta_0/\pi) \sigma^2$.