For a zero order reaction ,k= rate constant

When $t= t_{y_2},$ then $x= \large\frac{|A|_o}{2}$

Given that $[A_0]= 2M$

$t_{y_2}=1 hr,K=1$

$\therefore K= \large\frac{|A|_0}{2 \times ty_2} =\frac{2}{2 \times 1} $

$\qquad= 1 mol .L^{-1}.hr^{-1}$

Now, change in conc $(x)=0.50 -0.25$

$\qquad= 0.25\;M$

$\therefore t= \large\frac{x}{k} =\frac{0.25}{1} $$=0.25\;hr$

Hence c is the correct answer.