In the given figure, the emf of the cell is 2 . 2   V and if internal resistance is 0 . 6   Ω…

The above circuit in the question is arranged like that, so all the resistance are connected in parallel and the net resistance of the parallel resistor becomes, 

$$\begin{gathered} \frac{1}{R_P}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}=\frac{1}{4}+\frac{1}{6}+\frac{1}{12}+\frac{1}{8} \\ \Rightarrow R_P=\frac{48}{30} \mathrm{\Omega } \end{gathered}$$

Now the net resistance will be, $R_{net}=\left(\frac{48}{30}+0.6\right) \mathrm{\Omega }$, so the net power, $P=\frac{V^2}{R}=\frac{(2.2{)}^2}{2.2}=2.2 \mathrm{W}$