In the following circuit, the battery has an emf of 2V and an...

As we can see, two resistances of $2Ω$ in the middle portion(in both left and right side) are in parallel, so their combined resistance will be $\dfrac{2 \times 2}{2 + 2} + \dfrac{2 \times 2}{2 + 2} = 2Ω$. Now this resistance along with the topmost & bottommost resistance are in parallel again, so total resistance of the circuit will be,

${R}_{eq} = \dfrac{2}{3} + \dfrac{2}{3} = \dfrac{4}{3}Ω$

$∴ P = \dfrac{{V}^{2}}{{R}_{eq}} = \dfrac{4}{\dfrac{4}{3}} = 3W$