Answer
$1.41\;\rm kV$
Work Step by Step
We know that the energy stored in a capacitor is given by
$$U_C=\frac{1}{2}C(\Delta V_C)^2$$
Hence,
$$\Delta V_C=\sqrt{\dfrac{2U_C}{C}}$$
$$\Delta V_C=\sqrt{\dfrac{2 (1)}{(1\times 10^{-6}) }}$$
$$\Delta V_C=\color{red}{\bf 1.41}\;\rm kV$$
