Let [tex]B_i[/tex] denote the value on the [tex]i[/tex]-th drawn ball. We want to find the expectation of [tex]S=B_1+B_2+B_3+B_4+B_5[/tex], which by linearity of expectation is
[tex]E[S]=E\left[\displaystyle\sum_{i=1}^5B_i\right]=\sum_{i=1}^5E[B_i][/tex]
(which is true regardless of whether the [tex]X_i[/tex] are independent!)
At any point, the value on any drawn ball is uniformly distributed between the integers from 1 to 10, so that each value has a 1/10 probability of getting drawn, i.e.
[tex]P(X_i=x)=\begin{cases}\frac1{10}&\text{for }x\in\{1,2,\ldots,10\}\\0&\text{otherwise}\end{cases}[/tex]
and so
[tex]E[X_i]=\displaystyle\sum_{i=1}^{10}x\,P(X_i=x)=\frac1{10}\frac{10(10+1)}2=5.5[/tex]
Then the expected value of the total is
[tex]E[S]=5(5.5)=\boxed{27.5}[/tex]