$3 \mathrm{\Omega }$, $6 \mathrm{\Omega }$ and $2 \mathrm{\Omega }$ resistances are in parallel. So, the potential drop across them will be equal. Of these, three resistances maximum heat will be generated across $2 \mathrm{\Omega }$ resistance

        $\left(H=\frac{V^2}{R}t or H\propto \frac{1}{R}\right)$

Similarly, $5 \mathrm{\Omega }$ and $4 \mathrm{\Omega }$ resistances are also in parallel so, more heat will be generated across $4 \mathrm{\Omega }$ resistor. Now the given circuit can be redrawn as :


          


          $\frac{V_1}{V_2}=\frac{1}{{\displaystyle \frac{20}{9}}}=\frac{9}{20}$

$\therefore V_1=\left(\frac{9}{29}\right)V$

and    $V_2=\left(\frac{20}{29}\right)V$

Power developed across $2 \mathrm{\Omega }$ resistor will be

          $P_1=\frac{V_1^2}{2}={\left(\frac{9}{29}\right)}^2\frac{V}{2}$

and power developed across $4 \mathrm{\Omega }$ resistor is

          $P_2=\frac{V_2^2}{2}={\left(\frac{20}{29}\right)}^2\frac{V}{4}>P_1$

$\therefore$ Maximum heat is generated across $4 \mathrm{\Omega }$ resistance.