Use Gauss's law

The electric flux through any closed surface is given by $\Phi = \frac{q_{\text{enclosed}}}{\varepsilon_0}$ where $q_{\text{enclosed}}$ is the net charge inside the surface.

Identify which charges lie inside the sphere

The sphere has:

• center at origin
• radius $4a$

So any charge whose distance from the origin is less than $4a$ will be enclosed.

• Charge $5Q$ is at $(3a,0)$ $r_1 = 3a < 4a$ Hence, $5Q$ is inside the sphere.

• Charge $-2Q$ is at $(-5a,0)$ $r_2 = 5a > 4a$ Hence, $-2Q$ is outside the sphere.

Compute enclosed charge

Only $5Q$ is enclosed, so $q_{\text{enclosed}} = 5Q$

Compute flux

By Gauss's law, $\Phi = \frac{5Q}{\varepsilon_0}$

Match with options

$\boxed{\frac{5Q}{\varepsilon_0}}$ This corresponds to Option D.