\begin{equation*}
\mu = \int_{-1}^2 x \cdot x^2/3 dx = \int_{-1}^2 x^3/3 dx = \frac{2^4}{12} - \frac{(-1)^4}{12} = \frac{15}{12} = \frac{5}{4}
\end{equation*}
\begin{align*}
\sigma^2 & = E[X^2] - \mu^2\\
& = \int_{-1}^2 x^2 \cdot x^2/3 \; dx - \mu^2\\
& = \int_{-1}^2 x^4/3 \; dx - \left ( \frac{5}{4} \right )^2\\
& = \frac{2^5}{15} - \frac{(-1)^5}{15} - \frac{25}{16}\\
& = \frac{33}{15} - \frac{25}{16} = \frac{51}{80}
\end{align*}
which gives
\begin{equation*}
\sigma = \sqrt{\frac{51}{80}} \approx 0.7984.
\end{equation*}