Time required for concentration to change from $0.4$to $0.2$is same as the time required for the concentration to change from $0.2$to $0.1$, it means ${\text{t}}_{\text{1/2}}$is independent of initial concentration, i.e. it is a first order reaction.
And from the graph ${\text{t}}_{\text{1/2}}=20\text{s}$
$\therefore \text{K}=\frac{0.693}{20}{\text{s}}^{-1}$
$\therefore \text{Rate}=\text{k[A]}$
At $40$second $\text{[A]}=0.1$
$\therefore \text{rate}=\frac{0.693}{20}$
$=3.46\times {10}^{-3}{\text{Ms}}^{-1}$
$=3.5\times {10}^{-3}{\text{s}}^{-1}$