Width of central maxima is equal to $2 \times $ fringe width.

$ \qquad= 2 \times \beta$

$\qquad= 2 \large\frac{\lambda D}{d}$

$\qquad= \large\frac{2 \times 2.1 \times 5 \times 10^{-7}}{0.15 \times 10^{-2}}$

$\qquad= 1.4 \times 10^{-3} \;m$

$\qquad= 1.4 mm$

Hence b is the correct answer.