Given $|\overrightarrow{a}|=1$ and $|\overrightarrow{b}| =1$
$\theta$ is the angle between $\overrightarrow{a}$ and $\overrightarrow{b}$
$|\overrightarrow{a}+\overrightarrow{b}|=1$
$|\overrightarrow{a}+\overrightarrow{b}|^2=1$
$(\overrightarrow{a}+\overrightarrow{b})^2=1$
$\overrightarrow{a^2}+\overrightarrow{b^2}+2|\overrightarrow{a}||\overrightarrow{b}| \cos \theta=1$
$\overrightarrow{|a|^2}+\overrightarrow{|b|^2}+2|\overrightarrow{a}||\overrightarrow{b}| \cos \theta=1$
$1^2+1^2+2 \times 1\times 1 \cos \theta=1$
$\cos \theta = -\large\frac{1}{2}$
$\theta$ lies in the second quadrant
$\theta=\large\frac{2\pi}{3}$
Hence 4 is the correct answer.