Clearly, 2$\Omega$ , 4$\Omega$ and ( 1 + 5) $\Omega$ resistors are in parallel. Hence, potential difference is same across each of them.

$\therefore$ I₁ × 2 = I₂ × 4 = I₃ × 6

Given I₁ = 3A    $\therefore$ I₁ × 2 = I₃ × 6

Given I₁ = 3A.

$\therefore$ I₁ × 2 = I₃ × 6 provides

${I_3} = {{{I_1} \times 2} \over 6} = {{3 \times 2} \over 6} = 1A$

Now, the potential across the 5$\Omega$ resistor is
V = I₃ × 5 = 1 × 5 = 5V.

$\therefore$ the power dissipated in the 5$\Omega$ resistor

$P = {{{V^2}} \over R} = {{{5^2}} \over 5} = 5\,watt.$