$\frac{1}{R}=\frac{1}{4}+\frac{1}{6}+\frac{1}{12} \Rightarrow R=\frac{12}{6}=2 \,\Omega$

Again $R$ is connectedto $1.5 \,\mathrm{V}$ battery whose internal resistance $r=1\, \Omega$

Equivalent resistance now, $R^{\prime}=2\, \Omega+1 \,\Omega=3\, \Omega$

Current, $I_{\text {total }}=\frac{V}{R^{\prime}}=\frac{1.5}{3}=\frac{1}{2} \,\mathrm{A}$

$I_{\text {total }}=\frac{1}{2}=3 x+2 x+x=6 x$

$\Rightarrow x=\frac{1}{12}$

$\therefore $ Current through $4\, \Omega$ resistor $=3 x$

$=3 \times \frac{1}{12}=\frac{1}{4}\, A$

Therefore, rate of Joule heating in the $4 \,\Omega$ resistor

$=I^{2} R=\left(\frac{1}{4}\right)^{2} \times 4=\frac{1}{4}=0.25 \,\mathrm{W}$