The correct answer is $\boxed{{1 \over {{T_0}}}\sum\limits_{n = – \infty }^\infty {\exp } \left( {{{j\pi nt} \over {{T_0}}}} \right)}$.
The Fourier series of an impulse train is a sum of sine waves with frequencies that are integer multiples of the fundamental frequency. The fundamental frequency is equal to the reciprocal of the period of the impulse train. In this case, the period of the impulse train is $T_0$, so the fundamental frequency is $1/T_0$.
The Fourier series coefficients of an impulse train are given by $a_n = 1/T_0$ for all $n$. This means that the Fourier series of an impulse train is given by
$$s(t) = \frac{1}{T_0} \sum_{n=-\infty}^{\infty} \exp \left( j\frac{\pi n t}{T_0} \right)$$
This is the answer choice $\boxed{{1 \over {{T_0}}}\sum\limits_{n = – \infty }^\infty {\exp } \left( {{{j\pi nt} \over {{T_0}}}} \right)}$.
The other answer choices are incorrect because they do not include the factor of $1/T_0$ in the Fourier series coefficients. This factor is necessary to ensure that the Fourier series converges to the impulse train.