The DC circuit shown it the figure has a power source (battery), some resistors, switches and voltage and current meters. The switch S1 is open, so the right branch of the circuit is neglected. Assuming all meters are ideal (no resistive losses), we have a simple series circuit made of resistors R2 and R3.
We are required to compute the total power dissipated in the circuit, which corresponds to the total power delivered by the voltage source.
The power source of V= 6V is delivering a current in the series circuit equal to the current through R1 and R3. That current is the same as that read by M1 known to be I=0.04 A