Variance calculation from PSD
For a wide-sense-stationary random process, all the random variables
comprising the process have the same mean $\mu$ and variance $\sigma^2$, and
the variance is the integral of the power spectral density $S(f)$ less the square
of the mean:

$$\sigma^2 = \int_{-\infty}^{\infty} S(f)\,\,\mathrm df - \mu^2.$$

Generally, the mean is $0$ and $\sigma^2$ is just the integral of
$S(f)$, but when $\mu \neq 0$, then $S(f)$ includes
an impulse of $\mu^2\delta(f)$ at $f=0$ which contributes $\mu^2$ to
the integral, and this gets subtracted off by the $-\mu^2$ term in the
above formula. For discrete-time signals, the integrals should
be replaced by the appropriate sums.  But when there is a 
(deterministic) signal component and a (random, zero-mean)
noise component, calculation of the signal-to-noise ratio SNR
(which has many definitions including one which says it is
the ratio of the signal energy to the noise variance times
the signal duration), requires separating out
the two numbers needed from knowledge of their sum (which is
what the periodogram will give you). This is a lot trickier
and the answer will typically depends on details of the signal that you
have not shared with us.