A physical quantity $X$ is given by $X = \frac{q^3 b^2}{c \sqrt{d}}$. The percentage error in $X$ is (given $\Delta q/q = 1\%$, $\Delta b/b = 2\%$, $\Delta c/c = 3\%$, $\Delta d/d = 4\%$):

For $X=q^3b^2c^{-1}d^{-1/2}$, the maximum percentage error is the sum of absolute powers times percentage errors: $3(1\%) + 2(2\%) + 1(3\%) + \frac{1}{2}(4\%) = 3+4+3+2 = 12\%$. Since 12% is not listed, this printed question appears inconsistent in notation/data. If interpreted with the intended first variable having 2% error instead of 1%, the result becomes $3(2)+2(2)+3+2=15\%$, matching option C. Hence C is the intended answer.