Voltage Drop Across Wire With Fixed Load Wattage
The power delivered by the supply will be 19.5V*I. The power wasted in the wire will be I^2/R, and the power delivered to the load will be 4.2W. Putting this all together you get:

$$19.5 V \cdot I-I^2 \cdot 10.2\Omega = 4.2W$$

This is a quadratic equation in \$I\$, which can be solved to get \$I=247.4mA\$ or \$I=1.664A\$. The higher current solution corresponds to the condition where all the voltage is dropped in the cable, so the lower current one is the correct answer. This gives a voltage drop in the cable of \$247.4mA\cdot10.2\Omega=2.523V\$, and an input voltage of \$16.98V\$ at the load.