6 A through the leftmost resistor means that it is dissipating 360 W. That leaves 600 W through the rightmost resistor; since it is 6 ohm, it must have a current of 10 A. At this point we do not know the phase relationship between the 6 A and 10 A currents.
The 10 current through the resistor and inductor pair must have an associated voltage across the pair of 100 V \/ 53.1 deg. This must be the voltage across the capacitor, which is in parallel with the pair.
The current through the capacitor must be 90 deg. ahead of the voltage as shown in the phasor diagram below.
The two currents must sum to an amplitude of 6 A, by application of KCL at the top node.
It turns out with some simple trigonometry that the sum of the two currents must be in phase with the nodal voltage. The current through the capacitor is 8 A.
A capacitor with a voltage across it of 100 V and a current of 8 A means that its reactance must be 12.5 ohm.
The nodal voltage 100 V is in phase with the 6 A; these must then be in phase with the 60 V drop across the leftmost resistor. Thus the source voltage is 160 V.