Answer
$22\rm \;\mu F$
Work Step by Step
We know that the power is given by
$$P_w=\dfrac{\Delta E}{t}$$
where $\Delta E$ is the energy used during this process of discharging which is the energy stored in the capacitor.
Hence,
$$P_w=\dfrac{U}{t}$$
where $U=\dfrac{1}{2}C(\Delta V_C)^2$
$$P_w=\dfrac{C(\Delta V_C)^2}{2t}$$
Hence,
$$C=\dfrac{2tP_w}{(\Delta V_C)^2}$$
Plug the known;
$$C=\dfrac{2(10\times 10^{-6})(10)}{(3)^2}$$
$$C=2.22\times 10^{-5}\;\rm F=\color{red}{\bf 22.2}\;\mu F$$
