Calculate the flux

The flux through curved surface of cone B is same as the flux through the base, as cone B encloses no charge. Hence required flux is given by:

$\phi_{E} = \dfrac{20 \epsilon_{0}}{2 \epsilon_{0}} \bigg(1 - \cos \bigg((74/2)^{\circ}\bigg)\bigg) \approx 2 N m^2/C$

Note: The electric flux through a disk due to the E field of a point charge $q$ kept on its axis, such that the cone formed by the point and circumference has half angle $\theta$ is given by:

$\phi_{E} = \dfrac{q}{2 \epsilon_{0}} \bigg(1 - \cos \theta\bigg)$

for a disc electric flux through its surface can be easily derived by taking a ring element and calculating flux through it using $d \phi=E.dA$ and then integrating it for the whole disc $\frac{ q }{ 2 \epsilon } (1-\cos \theta)$ and one can even appreciate on the use of this formula is that we can find out the flux through an imaginary sheet by putting $\theta=90$

total flux emanating from the source in entire of 4pi steredians = 20, the solid angle of the cone A = 2pi(1-cos(semi-vertical angle)) = 2pi(1-cos(37)) = 2pi/5..., therefore since flux for 4pi steridians = 20, flux for 2pi/5 radians = 2