Answer
$\mathop \smallint \limits_{C'}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 44$
Work Step by Step
First, we find $\mathop \smallint \limits_{{C_2}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}}$, where ${C_2}$ is the path in counterclockwise direction as is shown in Figure 16.
We have
$\mathop \smallint \limits_C^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 5 = \mathop \smallint \limits_{{C_1}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} + \mathop \smallint \limits_{{C_2}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} + \mathop \smallint \limits_{{C_3}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}}$
Since $\mathop \smallint \limits_{{C_1}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 8$, $\mathop \smallint \limits_{{C_3}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 8$, so
$5 = 8 + \mathop \smallint \limits_{{C_2}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} + 8$
$\mathop \smallint \limits_{{C_2}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = - 11$
Thus, the value of $\mathop \smallint \limits_{C'}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}}$, where $C'$ is the path that traverses the loop ${C_2}$ four times in the clockwise direction is
$\mathop \smallint \limits_{C'}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 4\cdot\mathop \smallint \limits_{ - {C_2}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 4\cdot\left( { - \mathop \smallint \limits_{{C_2}}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}}} \right) = 4\cdot11 = 44$
So, $\mathop \smallint \limits_{C'}^{} {\bf{F}}\cdot{\rm{d}}{\bf{r}} = 44$.