Answer
${\bf 44.1}\;\rm Hz$
Work Step by Step
Recalling that the voltage across a discharging capacitor is given by
$$V_C=V_0e^{-t/RC}$$
And we know that $V_C=\frac{1}{2}V_0$ at $t_1=2.5$ ms, so
$$\frac{1}{2} \color{red}{\bf\not} V_0= \color{red}{\bf\not} V_0e^{-t/RC}$$
$$\ln\left[\frac{1}{2}\right]=\dfrac{-t_1}{RC}$$
Hence,
$$RC=\dfrac{-t_1}{\ln\left[\frac{1}{2}\right]}\tag 1$$
The crossover frequency is given by
$$f_c=\dfrac{\omega_c}{2\pi}$$
where $\omega_c=1/RC$;
$$f_c=\dfrac{1}{2\pi RC}$$
Plug from (1),
$$f_c=\dfrac{-\ln\left[\frac{1}{2}\right]}{2\pi t_1}$$
Plug the known;
$$f_c=\dfrac{-\ln\left[\frac{1}{2}\right]}{2\pi (2.5\times 10^{-3})}$$
$$f_c=\color{red}{\bf 44.1}\;\rm Hz$$
