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Power (P) = I^2R
I = Current, R = Resistance
R = rho*l/A
rho = Resistivity, l = Length, A = Cross sectional Area
Now l1= 0.9l
R1 = rho*l1/A = rho*0.9l/A = 0.9R
New Power (P1) = I^2R1 = I^2*0.9R = 0.9P
Hence the Power will also reduce by 10%.
Thanks. answer updated
Decrease by 10%. D must be the answer.
thanks for correction