Power absorbed by a resistor in a crossed circuit
The inductors are crossed.

The circuit is also symmetrical - such that the voltage across the top capacitor is opposite to the voltage of the bottom capacitor.  Similar rule applies to inductor current.  Let's exploit the symmetry.

Result:  the rms dissipated power in the resistor is indeed 2 W.

KCL:

\$ i_3=i_1+i_2  \$

KVL1: (around the voltage source, the two capacitors, and the resistor)

\$ V=2Z_1i_1+i_3R = 2Z_1i_1+Ri_1+Ri_2  \$

KVL2: (around the voltage source, L2, and C2)

\$ V=-Z_2i_2+i_1Z_1  \$

Solution:

\$ R=1\Omega, V = 2V, Z1 = 1/(\omega Cj) = -0.5j, Z2 = \omega Lj = 2j\$

\$ i_1= (V+VR/Z_2)/(2Z_1+R+RZ_1/Z_2)\$

\$ i_2= (i_1Z_1-V)/Z_2\$

\$ i_1= 1.6+0.8j\$

\$ i_2= -0.4+0.8j\$

\$ i_3= 1.2+1.6j\$

Absolute value of the resistor current phasor is 2 A.  The rms is then \$ \sqrt 2A \$ and the power dissipation \$ P = Ri_3^2=1\Omega*(\sqrt2A)^2=2W\$

Notice that due to the capacitor/resistor value symmetry the dissipated power is independent of the excitation frequency.

Schematic (Z1 are the capacitors, Z2 are the inductors):



Results:




  Thoughts on using the Thevenin equivalent circuit to solve this problem?


Not sure what Thevenin will give you here- the circuit can be solved using two KVLs! Also, the impedances are frequency dependent and Thevenin/Norton theorems are typically used for DC voltage/current sources and resistive networks.

Btw. thank you for this interesting question!  I had fun solving it :).