$\overline{){{\mathbf{K}}}_{\mathbf{a}\mathbf{}}{\mathbf{\times}}{\mathbf{}}{{\mathbf{K}}}_{{\mathbf{b}}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{{\mathbf{K}}}_{{\mathbf{w}}}}\phantom{\rule{0ex}{0ex}}{\mathbf{K}}_{\mathbf{a}}\mathbf{}\mathbf{=}\frac{{\mathbf{K}}_{\mathbf{w}}}{{\mathbf{K}}_{\mathbf{b}}}$

Given that Kb for C6H5NH3+ is 1.7 x 10^{-9} at 25 °C, what is the value of Ka for C6H5NH3+ at 25 °C? Ka=____?