= Solution
At unchanged intrinsic catalyst kinetics, increasing the electrochemically accessible true area by $100$ multiplies exchange current per geometric area by $100$. Equivalently, the local <current density> at a fixed total geometric current falls by $100$. Each <Tafel equation> activation loss decreases by $b\log_{10}100=2b$:
$$
\eta_{\rm H,new}=0.1475-0.0590=0.0885\ \mathrm V,
\qquad
\eta_{\rm O,new}=0.7200-0.1800=0.5400\ \mathrm V.
$$
Thus \b[$V_{\rm new}=0.59948\ \mathrm V$ and $P_{\rm new}/A=5.99\ \mathrm{W\,m^{-2}}$], an increase of $2.39\ \mathrm{W\,m^{-2}}$, or about $66.3\%$. Power does not increase a hundredfold, because the voltage improvement is logarithmic and the specified geometric current is unchanged. This assumes both <electrodes> gain accessible active area, the original exchange currents use the original area basis, and the porous structure does not introduce additional transport or Ohmic losses.
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