$[N_2O_5]$ left $=\,1.0-0.2\,=\,0.8\,mol\,L^{-1}$
Rate of reaction $=\,k\times [N_2O_5]$
$=\,3.0\times 10^{-4}\times 0.8$
$=\,2.4\times 10^{-4}\,mol\,L^{-1}\,s^{-1}$
Rate of formation of $NO_2$
$=\,4\times 2.4\times 10^{-4}\,=\,9.6\times 10^{-4}\,mol\,L^{-1}\,s^{-1}$
$(\log \,4 = 0.60,\, \log \,5 = 0.69)$