Power depends on voltage across the circuit and resistance of the circuit
\begin{equation} P = \frac{V^2}{R};\\ P_{series} = \frac{V^2}{(R_1+R_2)};\\ P_{parallel} = \frac{V^2}{(R_1^{-1}+R_2^{-1})}=\frac{V^2}{\frac{R_1+R_2}{R_1R_2}}=\frac{V^2R_1R_2}{R_1+R_2};\\ \frac{P_{series}}{P_{parallel}} = \frac{1}{R_1R_2} \end{equation}
i.e. the answer depends on the product of the two resistances.
i.e. if $R_1R_2 > 1$, $P_{series} < P_{parallel}$
if $R_1R_2 < 1$, $P_{series} > P_{parallel}$