We have,

$P = {{{V^2}} \over R}$

$\Rightarrow 36 = {{{V^2}} \over 9}$

$\Rightarrow V = 18$

Current passing through the 9$\Omega$ resistor is

${i_1} = {V \over R} = {{18} \over 9} = 2A$

The 9$\Omega$ and 6$\Omega$ resistors are in parallel, therefore

${i_1} = {6 \over {9 + 6}} \times i$

where i is the current delivered by the battery.

$\therefore i = {{2 \times 15} \over 6} = 5A$

Thus, potential difference across 2$\Omega$ resistor is

V = iR = 5 × 2 = 10V