The equivalent circuit of the given circuit will be
From circuit, $\frac{1}{R} =\frac{1}{3}+\frac{1}{6}$
$=\frac{2+1}{6}=\frac{3}{6} \,\Omega$
$R_1=2\,\Omega$
$R_{2} =R_{2}+R_{2}+R_{4} $
$=2+2+4=8 \,\Omega$
The internal resistance of battery $=1 \,\Omega$
So, the equivalent resistance of circuit
$=8 \Omega+1 \Omega $
$=9 \,\Omega$
The current in the circuit
$I=\frac{V}{R}=\frac{45}{9}$
$=\frac{1}{2} A$
The power dissipated in $3 \Omega$ resistance
$P =I^{2} \times R$
$=\left(\frac{1}{2}\right)^{2} \times 3 $
$=\frac{1}{4} \times 3 $
$=0.75 \,W$
