Correct option is A. $$8.3 \times 10^{-10} C$$
The electric field between two close parallel plates is,
$$E=\dfrac{\sigma}{\epsilon_0}$$
surface charge density is
$$\sigma = Q/A$$
$$E=\dfrac{\dfrac{Q}{A}}{\epsilon_0}$$$$=\dfrac{Q}{\epsilon_0A}$$
$$Q=\epsilon_0EA$$
$$=\epsilon_0El^2$$
$$Q=(8.85\times 10^{-12}C^2N.m^2)(130N/C)(0.85m)^2$$
$$=8.3\times 10^{-10}C$$