We want the variance of $X_1+\cdots+X_{10}$. The calculation will be done "by hand," using basic ideas only. Recall the useful formula $\text{Var}(Y)=E(Y^2)-(E(Y))^2.$ This is not difficult to derive from the usual official definition $\text{Var}(Y)=E(Y-\mu_Y)^2$, and is often easier to compute with.
The expectation of $X_1+\cdots +X_{10}$ is easy, for in general the expectation of a sum is the sum of the expectations. It follows that $E(X_1+\cdots+X_{10})=\frac{50}{3}.$
The expectation of $(X_1+\cdots+X_{10})^2$ is more complicated. Expand the square. We get $\sum_{i=1}^{10}X_i^2 +2\sum_{1\le i We want the expectation of the complicated expression above. First, we compute $E(X_i^2)$. This random variable takes on values $1^2$, $2^2$, and $3^2$ with probabilities respectively equal to $\frac{15}{30}$, $\frac{10}{30}$, and $\frac{5}{30}$. It follows that $E(X_i^2)=\frac{15}{30}\cdot 1 +\frac{10}{30}\cdot 4+\frac{5}{30}\cdot 9.$ This simplifies to $\frac{10}{3}$. It follows that $\sum_{i=1}^{10}E(X_i^2)=\frac{100}{3}.$ Finally, we need to evaluate the expectation of a typical "mixed" term $X_iX_j$. This requires some calculation. If $X_i=1$, (probability $\frac{15}{30}$), $X_j$ is $1$, $2$, and $3$ with probabilities respectively equal to $\frac{14}{29}$, $\frac{10}{29}$, and $\frac{5}{29}$. Thus we get a contribution to $E(X_iX_j)$ equal to
$\frac{15}{30}\left(\frac{14}{29}\cdot 1+\frac{10}{29}\cdot 2+\frac{5}{29}\cdot 3\right)$.
When $X_i=2$ (probability $\frac{10}{30}$), we get contribution to $E(X_iX_j)$ equal to
$\frac{10}{30}\left(\frac{15}{29}\cdot 2+\frac{9}{29}\cdot 4+\frac{5}{29}\cdot 6\right)$.
Finally, when $X_i=3$, the contribution to the expectation is
$\frac{55}{30}\left(\frac{15}{29}\cdot 3+\frac{10}{29}\cdot 6+\frac{4}{29}\cdot 9\right)$.
Add up all these terms. After simplifying, we get $E(X_iX_j)=\frac{80}{29}.$ There are $\binom{10}{2}$ terms $X_1X_j$. It follows that $E(2\sum_{1\le i Finally, we put the various pieces together: $\text{Var}\left(\sum_{i=1}^{10} X_i\right)=\frac{100}{3}+\frac{7200}{29}-\left(\frac{50}{3}\right)^2.$ This turns out to be approximately $3.83$.
For the variance of $\frac{1}{10}(X_1+\cdots +X_{10})$, we can use the general fact that $\text{Var}(kY)=k^2\text{Var}(Y)$ for any constant $k$.