Correct option is A. $$\dfrac{nd}{n-1}$$
Total internal reflection takes place for all angles greater than a critical angle. Thus, for the smallest radius of curvature, the angle made at the right end of the fibre $$(i)$$ should be equal to the critical angle, for total internal reflection. Thus,
$$\dfrac{\sin90^{\circ}}{\sin i} = \dfrac{n_1}{n_2}$$
But, $$n_2 = 1$$, and $$n_1 = n$$
$$\Rightarrow \sin i = \dfrac{1}{n}$$
$$\Rightarrow \dfrac{r-d}{r} = \dfrac{1}{n}$$
$$\Rightarrow n(r-d) = r$$
$$\Rightarrow nr - r = nd$$
$$\Rightarrow r(n-1) = nd$$
$$\therefore r = \dfrac{nd}{n-1}$$