Answer
${\bf 1.59}\;\rm \mu F$
Work Step by Step
The crossover frequency in a simple RC circuit is at $V_R=V_C$, where the crossover frequency is given by
$$\omega_c=2\pi f_c$$
So,
$$f_c=\dfrac{\omega_c}{2\pi}$$
where $\omega_c=1/RC$, so
$$f_c=\dfrac{1}{2\pi RC}$$
So the capacitance is given by
$$C=\dfrac{1}{2\pi Rf_c}$$
Plug the given;
$$C=\dfrac{1}{2\pi (1000)(100)}=\color{red}{\bf 1.59}\;\rm \mu F$$
