Explanation

(b) $E = \frac{L^{2}}{2I}\ \therefore E \propto L^{2} \Rightarrow \frac{E_{2}}{E_{1}} = \left( \frac{L_{2}}{L_{1}} \right)^{2}$
$\frac{E_{2}}{E_{1}} = \left\lbrack \frac{L_{1} + 200\%\ of\ L_{1}}{L_{1}} \right\rbrack = \left\lbrack \frac{L_{1} + 2L_{1}}{L_{1}} \right\rbrack^{2} = (3)^{2} \Rightarrow E_{2} = 9E_{1}$
Increment in kinetic energy $\mathrm{\Delta}E = E_{2} - E_{1} = 9E_{1} - E_{1}$
$\mathrm{\Delta}E = 8E_{1}\ \therefore\frac{\mathrm{\Delta}E}{E_{1}} = 8$ or percentage increase = 800%