Correct option is C. time taken by the particle to hit T could be $$\sqrt{\dfrac{5}{6}}\mu s$$ us as well as $$\sqrt{\dfrac{5}{6}}\mu s$$
$$a=10^{10}\times 400\sqrt{3}$$
$$a=4\sqrt{3}\times 10^{12}m/s^2$$
$$R=\dfrac{4^2\sin 2\theta}{a}$$
$$\dfrac{5\times 4\sqrt{3}\times 10^{12}}{4\times 10\times 10^{12}}=\sin 2\theta$$
$$\dfrac{\sqrt{3}}{2}=\sin 2\theta$$
$$2\theta=60^0,120^0$$
$$\theta=30^0$$ or $$60^0$$ for same range
$$T=$$$$\dfrac{2u\sin\theta}{a}$$$$=\dfrac{2\times 2\sqrt{10}\times 10^6}{2\times 4\sqrt{3}\times 10^{12}}$$
$$T=\sqrt{\dfrac{10}{12}}\times 10^{-6}$$
$$T=\sqrt{\dfrac{5}{6}}\times 10^{-6}$$