The rubber chord has a force constant k given by $k=\large\frac{yA}{L}$
$\qquad= \large\frac{5.0 \times 10^8 \times 1.0 \times 10^{-6}}{0.1}$
$\qquad= 5.0 \times 10^3 N/m$
Now from conservation of energy theorem.
The elastic potential energy of string is equal to the kinetic energy of the projectile
$\large\frac{1}{2} $$k ( \Delta l)^2=\large\frac{1}{2}$$mv^2$
Where $\Delta L$ is the extension in the string
$\therefore v=\sqrt {\large\frac{k}{m}}$$\Delta l$
$\qquad= \sqrt {\frac{5.0 \times 10^3}{5.0 \times 10^{-3}}}$$ 2 \times 10^{-2}$
$\qquad=20 m/s$
Hence b is the correct answer.