Correct option is A. $$13.5\times 10^{11}N$$
$$q_{1}=3\times 10^{-5}$$ $$q_{2}=5\times 10^{4}$$
$$r=10\times 10^{-2}m$$
$$F_{e}=\dfrac{1}{4\pi \varepsilon_{0}}\dfrac{q_{1}q_{2}}{r^{2}}$$
$$\dfrac{9\times 10^{9}\times 3\times 10^{-5}\times 5\times 10^{4}}{(10\times 10^{-2})^{2}}$$
$$\dfrac{9\times 15\times 10}{100\times 10^{-4}}$$
$$\dfrac{135}{100}\times 10^{12}$$
$$=1.35\times 10^{12}$$
$$=13.5\times 10^{11}$$