Rate constant of first order reaction
$\hspace10mm k=\frac{2.303}{t} \, log_{10} \frac{(A)_0}{(A)_t}$
or $\hspace10mm k=\frac{2.303}{1} \times log_{10} \, \frac{0.8}{0.2} \hspace15mm ...(i)$
(because 0.6 mole of B is formed)
Suppose $t_1$ hour are required for changing the concentration of A from 0.9 mole to 0.675 mole of B.
Remaining mole of $A = 0.9 - 0.675 = 0.225$
$\therefore \hspace10mm k=\frac{2.303}{t_1} \, log_{10} \, \frac{0.9}{0.225} \hspace15mm ...(ii)$
From Eqs. (i) and (ii)
$ \, \, \, \, \, \, \frac{2.303}{1} \, log_{10} \, \frac{0.8}{0.2}=\frac{2.303}{t_1} log_{10} \, \frac{0.9}{0.225}$
$ \, \, \, \, \, \, \, 2.303 \, log_{10} \, 4=\frac{2.303}{t_1} \, log_{10} \, 4$
$\hspace25mm t_1=1 \, h$