$R_{1}=3+1=4 \,\Omega, R_{2}=8\, \Omega$

Let $i$ be the total current in the circuit.

Current through $R_{1}$ is $i_{1}=\frac{i \times R_{2}}{R_{1}+R_{2}}=\frac{i \times 8}{12}=\frac{2 i}{3}$

Current through $R_{2}$ is $i_{2}=\frac{i \times R_{1}}{R_{1}+R_{2}}=\frac{i \times 4}{12}=\frac{i}{3}$

Power dissipated in $3\, \Omega$ resistor is

$P_{1}=i_{1}^{2} \times 3$       ....$(i)$

Power dissipated in $8 \Omega$ resistor is

$P_{2}=i_{2}^{2} \times 8$       ....$(ii)$

$\therefore \quad \frac{P_{1}}{P_{2}}=\frac{i_{1}^{2} \times 3}{i_{2}^{2} \times 8}$ or, $\frac{P_{1}}{P_{2}}=\frac{(2 i / 3)^{2} \times 3}{(i / 3)^{2} \times 8}=\frac{12}{8}=\frac{3}{2}$

$P_{1}=\frac{3}{2} \times P_{2}=\frac{3}{2} \times 2=3 \,\text { watt }$

$\therefore$ Power dissipated across $3 \Omega$ resistor is $3\,watt.$