The passage of electricity through dil ${\mathrm{H}}_2{\mathrm{SO}}_4$ for 16 minutes liberates a total of 224 ml of ${\mathrm{H}}_2$.The strength of the current in amperes will be

${\mathrm{H}}_2$ liberates 224 ml for time 16 mins
t=16x60sec,
We have to find out strength of the current i=?
If weight of 22400 ml of ${\mathrm{H}}_2$ is 2 gram then , 

 $\mathrm{Weight} \mathrm{of} 224 \mathrm{ml} \mathrm{of} {\mathrm{H}}_2=\frac{2}{100}=\frac{1}{50}\mathrm{gram}$
Now we have ${\mathrm{W}}_{{\mathrm{H}}_2}=\frac{1}{50}, {\mathrm{E}}_{{\mathrm{H}}_2}=1\mathrm{g}, \mathrm{E}=1 (2{\mathrm{H}}^{+}+2{\mathrm{e}}^{-}\to {\mathrm{H}}_{2 , }\mathrm{E}=\frac{2}{2}=1)$
According to Faradays first law.
$\mathrm{W}=\frac{\mathrm{E}\times \mathrm{i}\times \mathrm{t}}{96500}$
By putting all above values in equation
$$\begin{gathered} \frac{1}{50}=\frac{1\times \mathrm{i}\times 16\times 60}{96500} \\ \mathrm{i}=\frac{965}{5\times 16\times 6}=2.01\mathrm{A} \end{gathered}$$

Hence option D is correct.