Find the answer for this?

Let the voltage between $A$ and $B$ be $V_{AB}$ and the voltage across $R$ be $V$ . Then the power generated by $R$ is given by:

$\begin{aligned} P & = \frac {V^2}R = \left(\frac {R_2||R}{R_1+R_2||R}V_{AB}\right)^2 \times \frac 1R = \left(\frac {30||R}{20+30||R}\right)^2 \frac {V_{AB}^2}R \\ & = \left(\frac {\frac {30R}{R+30}}{20+\frac {30R}{R+30}}\right)^2 \frac {V_{AB}^2}R = \frac {9RV_{AB}^2}{(5R+60)^2} \end{aligned}$

Let us find the change of $P$ due to a change in $R$ .

$\begin{aligned} \frac {dP}{dR} & = \frac {9(5R+60)-2\cdot 9R(5)}{(5R+60)32} =\frac {540-45R}{(5R+60)^3} \end{aligned}$

$\implies \dfrac {dP}{dR} = 0$ , when $45R=540$ $\implies R = \boxed {12}\ \Omega$ .