In the circuit shown, the power developed across the 10 kilo ohm resistor is

Yeah this one's tricky, the V squared over R rule does work but only if you use the voltage across that resistor, not the full 200 V. The two resistors are in series so the 200 V splits between them, 50 V over the 10 k and 150 V over the 30 k, and 50 squared divided by 10000 gives 0.25 W, which is the 250 mW option. Think of it like water pressure dropping along a hose, the bit of pressure lost over the first section is what does the work there, not the pressure back at the tap. Easier still is to find the current first, 200 V over 40 k is 5 mA, then P equals I squared R gives the same 250 mW.