For the circuit shown in the figure

Total resistance of the circuit,

$R=2+\frac{6\times 1.5}{6+1.5}=3.2k\Omega$

Now   $I=\frac{24}{3.2\times {10}^3}=7.5\times {10}^{-3}A$    Option (a) is correct.

Current in $R_L$, $i_1=\frac{7.5\times {10}^{-3}\times 6}{\left(6+1.5\right)}=6\times {10}^{-3}A$

Potential difference across $R_L$ is $V_L=i_L{R}_L=6\times {10}^{-3}\times 1.5\times {10}^3=9\text{V}$. Option (b) is wrong.

Power through load, $P_1={i_L}^2\times R_L={\left(6\times {10}^{-3}\right)}^2\times 1.5\times {10}^3$

$=54\times {10}^{-3}J$

$\frac{P_1}{P_2}=\frac{i_1^2{R}_1}{i_2^2{R}_2}=\frac{{\left(7.5\times {10}^{-3}\right)}^2\times \left(2\times {10}^{-3}\right)}{{\left(1.5\times {10}^{-3}\right)}^2\times \left(6\times {10}^{-3}\right)}=\frac{25}{3}$   Option (c) is wrong.

After $R_1 \mathrm{and} {\mathrm{R}}_2$ are interchanged, total resistance

$R=6+\frac{2\times 1.5}{2+1.5}=6.86k\Omega$

Total current, $I=\frac{24}{6.86\times {10}^3}=3.5\times {10}^{-3}A$

Now current in $R_L=\frac{3.5\times {10}^{-3}\times 2}{\left(2+1.5\right)}=2\times {10}^{-3}A$

Power through load, ${P_1}^{\text{'}}={\left(2\times {10}^{-3}\right)}^2\times 1.5\times {10}^3$

$=6\times {10}^{-3}J$

Clearly  $P_1:{P_1}^{\text{'}}=9$      Option (d) is correct.