Answer
$P_t$ = 22.5 W
Work Step by Step
The voltage of the battery is the voltage across the resistance $R_1$, so we use the given values of $P_1$ to get the voltage of the battery as next
\begin{gather*}
P_{1} = \dfrac{V^{2}}{R_{1}} \\
\end{gather*}
Solve for $V$
\begin{align*}
V = \sqrt{P_{1}R_{1}} = \sqrt{(36 \,\text{W})(25 \mathrm{~\Omega})} = 30 \,\text{V}
\end{align*}
After $R_{2}$ is connected, the voltage drop in the battery will be due to the equivalent resistance which is calculated by
$$ R_{eq} = R_{1} + R_{2} = 25 \mathrm{~\Omega} + 15 \mathrm{~\Omega} = 40 \mathrm{~\Omega} $$
The total dissipated power by the two resistors is calculated by
\begin{gather*}
P_{t} = \dfrac{V^{2}}{R_{eq}}
\end{gather*}
Substitute the values for $V, R_{1}$ and $R_{2}$to get the total dissipated power
\begin{align*}
P_{t} = \dfrac{V^{2}}{R_{eq}} = \dfrac{(30 \,\text{V})^{2}}{ 40 \mathrm{~\Omega} } = \boxed{22.5 \,\text{W}}
\end{align*}
