Denote $X_1, X_2, X_3$ be the number of times that the three balls with number 1, 2, and 3 respectively, are selected in the 10 random draws. Since with replacement, $X=(X_1, X_2, X_3)$ follows a multinomial distribution with probabilities $p_1 = 0.21, p_2 = 0.36, p_3 = 0.43$. Following the distribution, we have
$$\mathbf{E}(X_i) = np_i; \hspace{20pt} \mathbf{Var}(X_i) = np_i(1-p_i); \hspace{20pt} \mathbf{Cov}(X_i, X_j) = -np_ip_j ( i \neq j).$$
We then have $$\bar{X} = \frac{1*X_1 + 2*X_2 + 3*X_3} {n} = \frac{X_1 + 2X_2 + 3X_3}{10} $$
$$ \mathbf{E}(\bar{X}) = \frac{1}{10} \{ \mathbf{E}(X_1) + \mathbf{E}(2X_2) + \mathbf{E}(3X_3) \}=p_1 + 2p_2 + 3p_3$$ $$=0.21 + 2*0.36 + 3*0.43 = 2.22\\$$
Similarly, $$\mathbf{Var}(\bar{X}) = \frac{1}{100} \{\mathbf{Var}(X_1) + 4\mathbf{Var}(X_2) + 9\mathbf{Var}(X_3) +4\mathbf{Vov}(X_1, X_2) + 12\mathbf{Cov}(X_2, X_3) + 6\mathbf{Cov}(X_1, X_3)\}.$$
EDIT:
The $X_1, X_2, X_3$ are correlated, so there should be covariance parts. I just added those.
Note that $\mathbf{Var}(aX + bY) = a^2\mathbf{Var}(X) + b^2\mathbf{Var}(Y) + 2ab\mathbf{Cov}(X,Y).$
END EDIT.
You can do the simple calculation by yourself.
Hope it helps.
[self-study]
tag & read its wiki, & we can re-open this for you. $\endgroup$