In the conventional USLE slope-length relation,$L\propto\lambda^m$ and m=.5 for slopes above 5%. Halving length makes predicted loss $A_2/A_1=(1/2)^{.5}=.7071068$. Reduction is $100(1-.7071068)=\mathbf{29.29\%}$. This uses the conventional USLE exponent intended by the problem.
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