The correct answer is (a) \(\sum_{-\infty}^\infty x(nT) \frac{sin(π/T) (t-nT)}{(π/T)(t-nT)}\)
To explain: x(t) = \(\sum_{-\infty}^\infty x(nT) \frac{sin(π/T) (t-nT)}{(π/T)(t-nT)}\) where the sampling interval T = 1/Fs=1/2B, Fs is the sampling frequency and B is the bandwidth of the analog signal.