As we know that impulse is given by

$$I = \Delta P = F \times \Delta t$$

or $ I=$ Area of $f-t$ graph

For fig (a)

$\rightarrow \mathrm{I}=\frac{1}{2} \times$ base $\times$ height

$$ =\frac{1}{2} \times 0.5 \times 1=0.25 \mathrm{~N}-\mathrm{sec} . $$

For fig (b),

$I=$ length $\times$ width

$$ =2 \times 0.5=1 \mathrm{~N} \text {-sec } $$

For fig (c),

I $=\frac{1}{2} \times$ base $\times$ height

$$ =\frac{1}{2} \times 1 \times 0.75=0.375 \mathrm{~N} \text {-sec. } $$

For fig (d),

$I=\frac{1}{2} \times$ base $\times$ height

$$ =\frac{1}{2} \times 2 \times 0.5=0.5 \mathrm{~N}-\mathrm{sec} \text {. } $$

Impulse is highest for that figure, whose area under $F$-t is maximum and i.e. figure(b)

Option (D) is correct.