Flux and symmetry

The flux through the first one is given by $\frac { 1 }{ 2 } \left( \frac { \theta }{ 2\pi } \frac { { q }_{ 0 } }{ { \epsilon }_{ 0 } } \right) =\frac { { tan }^{ -1 }2\sqrt { 2 } }{ 4\pi } \frac { { q }_{ 0 } }{ { \epsilon }_{ 0 } }$ . The flux through the second one is given by $\frac { 1 }{ 3 } \left( \frac { 1 }{ 4 } \frac { { q }_{ 0 } }{ { \epsilon }_{ 0 } } \right)$ . Adding both gives the required answer.