Solution

Initially, there are charge $q=CV$ and energy $CV^2/2$ stored on the left capacitor. When the circuit is closed, the same current flows through both resistors and the ratio of the heat generated in each of them is

$$\frac{Q_1}{Q_2}=\frac{R_1}{R_2}.$$

When the circuit reaches steady state and there is no current, there are charge $q/2$ and energy $(q/2)^2/(2C)=CV^2/8$ stored on each capacitor. Thus, the total heat generated is

$$Q_1+Q_2=\frac{1}{2}CV^2-2\cdot\frac{1}{8}CV^2=\frac{1}{4}CV^2.$$

Solving both displayed equations for $Q_1$ and $Q_2$, we have

$$Q_1=\frac{1}{4}CV^2\cdot\frac{R_1}{R_1+R_2}\qquad\mbox{and}\qquad Q_2=\frac{1}{4}CV^2\cdot\frac{R_2}{R_1+R_2}.$$