Getting the standard deviation from the pdf 
A normally distributed random variable with mean $\mu$ has a probability density function given by $\dfrac{\gamma}{\sqrt{2\pi\sigma}}$ $\exp(-\dfrac{\gamma ^2}{\sigma} \dfrac{(x-\mu)^2}{2})  $

So the standard deviation is the square root of the variance, which is $E[(x-\mu)^2]$. However, I don't know how to proceed with this information. How can I get the standard deviation from the pdf?
 A: The probability density function of a normal random variable with the mean $\mu$ and the variance $\sigma^2$ is given by
$$
f(x, \mu, \sigma) = \frac{1}{\sigma \sqrt{2\pi} } e^{ -\frac{(x-\mu)^2}{2\sigma^2} }.
$$
Hence, the standard deviation of the normal random variable in your example is $\sqrt \sigma/\gamma$.
A: If $\rho$ denotes the standard deviation then  $$\frac{\gamma}{\sqrt{2\pi\sigma}}=\frac{1}{\rho\sqrt{2\pi}}$$
A: Usually one writes
$$
\frac1{\sigma\sqrt{2\pi}}\exp\left(-\frac 1 2 \left( \frac{x-\mu}\sigma \right)^2\right). \tag 1
$$
If one must add that extra parameter $\gamma$, then one has
$$
\frac\gamma{\sigma\sqrt{2\pi}}\exp\left(-\frac 1 2 \gamma^2\left( \frac{x-\mu}\sigma \right)^2\right). \tag 2
$$
This amounts to just putting $\sigma/\gamma$ where $\sigma$ had been.  If one can show that for $(1)$, the standard deviation is $\sigma$, then it follows that for $(2)$ the standard deviation is $\sigma/\gamma$. PS: I see the question has undergone further editing, having $\sqrt{\sigma}/\gamma$ where $\sigma$ appears in $(1)$.  In that case, the standard deviation would be $\sqrt{\sigma}/\gamma$. end of PS
The variance is
$$
\int_{-\infty}^\infty (x-\mu)^2 f(x)\,dx.
$$
In the case of $(1)$, this is
\begin{align}
& \int_{-\infty}^\infty (x-\mu)^2 \frac1{\sigma\sqrt{2\pi}}\exp\left(-\frac 1 2 \left( \frac{x-\mu}\sigma \right)^2\right)\,dx \\[10pt]
= {} & \sigma^2 \int_{-\infty}^\infty \left( \frac{x-\mu}\sigma \right)^2 \frac1{\sqrt{2\pi}}\exp\left(-\frac 1 2 \left( \frac{x-\mu}\sigma \right)^2\right)\,\frac{dx}\sigma \\[10pt]
= {} & \frac{\sigma^2}{\sqrt{2\pi}} \int_{-\infty}^\infty w^2 \exp\left(-\frac 1 2 w^2\right)\,dw.
\end{align}
So it is enough to show that without the $\sigma^2$ we get $1$.  Since we have an even function over an interval symmetric about $0$, the integral is
\begin{align}
& \frac2{\sqrt{2\pi}} \int_0^\infty w^2 \exp\left(-\frac 1 2 w^2\right)\,dw = \frac2{\sqrt{2\pi}} \int_0^\infty w \exp\left(-\frac 1 2 w^2\right)(w\,dw) \\[10pt]
= {} & \frac2{\sqrt{\pi}} \int_0^\infty \sqrt{u} e^{-u}\,du = \frac2{\sqrt{\pi}} \Gamma\left(\frac 3 2 \right) = \frac2{\sqrt{\pi}}\cdot\frac 1 2 \Gamma\left(\frac 1 2 \right).
\end{align}
Now recall that $\Gamma(1/2)= \sqrt{\pi}$.  As to how we know that, that is the topic of another question already posted here.
