Hello!
To determine [H₃O⁺], we need to apply the Henderson-Hasselback equation, since this is a case of an acid and its conjugate base:
[tex]pH=pKa+log( \frac{[A^{-}] }{[HA]} )[/tex]
[tex]pH=4,76+log( \frac{0,030M}{0,25M} ) \\ \\ pH=3,84[/tex]
Now, we use the definition of pH and clear [H₃O⁺] from there:
[tex]pH=-log[H_3O^{+}] [/tex]
[tex] [H_3O^{+}] = 10^{-pH} =10^{-3,84}=0,00014 M[/tex]
So, the [H₃O⁺] concentration is 0,00014 M
Have a nice day!