Correct choice is (a) \(\frac{ω_c}{\pi}.\frac{sin ω_c.n}{ω_c.n}\)
The best I can explain: We know that, x(n)=\(\frac{1}{2\pi} \int_{-\pi}^{\pi}X(\omega)e^{j\omega n} dω\)
=\(\frac{1}{2\pi} \int_{-ω_c}^{ω_c}1.e^{j\omega n} dω=\frac{sin ω_c.n}{ω_c.n}\)
=\(\frac{ω_c}{\pi}.\frac{sin ω_c.n}{ω_c.n}\)