Use Gauss's law

The electric flux through any closed surface is given by $\Phi = \frac{Q_{\text{enclosed}}}{\varepsilon_0}.$

Flux through cube $C_1$

From the figure/text, the inner cube $C_1$ encloses charge $2Q$. Hence, $\Phi_1 = \frac{2Q}{\varepsilon_0}.$

Flux through cube $C_2$

The outer cube $C_2$ encloses both charges, since it is concentric and larger. So total enclosed charge by $C_2$ is $2Q + 3Q = 5Q.$ Therefore, $\Phi_2 = \frac{5Q}{\varepsilon_0}.$

Find the ratio

$\Phi_1 : \Phi_2 = \frac{2Q}{\varepsilon_0} : \frac{5Q}{\varepsilon_0} = 2:5.$

Match with options

$2:5$ corresponds to Option C.

Compare with stored correct answer

Stored correct answer = C.

This matches our derived answer.