ANS:B - False

If assuming the voltage source to be a constant value. Here is an example why:

Suppose we have a simple series circuit with V = 50 V & R = 10k ohms,
Voltage rating using P= V^2/R =50^2/10k = 0.25 W,
Using P=I^2*R we need to get the value of current first, I =V/R=50/10k 5mA, then P =(5mA)^2(10k) = 0.25W,

But what if we increase the value of resistance, say we make it as 20k ohms with the same value of voltage.

P =V^2/R = 50^2/20k = 0.125W, Using P=I^2*R we again need to solve for the current first,
I = V/R = 50/20k.
I = 2.5mA (CURRENT DECREASES), then P=(2.5mA)^2(20k)=0.125W the power still decreases, why?

Because as the value of resistance increases with a constant voltage source the current also decreases thus, the formulas P=V^2/R and I^2*R will always give the same answer. So for P=V^2/R if R increases the P decreases, for P=I^2*R if R increases, the Current would decrease and Power would still decrease as shown in the example above.