Expanding on JohnK's comment: You want to find the expectation of a function with respect to a standard normal distribution. Since it seems difficult to evaluate the integral, we can instead resort to estimating it using Monte Carlo.
If we sample $x_1, x_2, \dots x_N$ from a $N(0,1)$ distribution, then your quantity can be estimated by
$$\dfrac{1}{n} \sum_{i=1}^{N} \sigma(x_i) (1 - \sigma(x_i)\,. $$
As you increase $N$, you can an increasingly better estimate of the quantity. The R code below returns the value .20669 with standard error .00015.
> set.seed(100)
> sig <- function(x)
+ {
+ 1/(1+exp(-x))
+ }
>
> N <- 1e5
> samples <- rnorm(N, 0, 1)
> mean(sig(samples)*(1 - sig(samples)))
[1] 0.2066863
> # The standard error of the mean estimate
> sd(sig(samples)*(1 - sig(samples)))/sqrt(N)
[1] 0.0001463474