First, we must find $E(X)$, which is
$$
E(X)=0 \cdot 0.3+1 \cdot 0.5+2 \cdot 0.1+3 \cdot 0.1=1
$$
Now we have
$$
\begin{gathered}
\sigma^{2}(X)=(0-1)^{2} \cdot 0.3+(1-1)^{2} \cdot 0.5+(2-1)^{2} \cdot 0.1+(3-1)^{2} \cdot 0.1 \\\\
=0.3+0+0.1+0.4=0.8
\end{gathered}
$$
Thus, the variance of $X$ is $0.8$.
