The number of molecules in 100 mL of 0.02 N ${\mathrm{H}}_2{\mathrm{SO}}_4$ is 

$\begin{array}{l}\mathrm{Given} , {\mathrm{N}}_{{\mathrm{H}}_2{\mathrm{SO}}_4}=0.02\mathrm{N} ; \mathrm{V} = 100\mathrm{ml}\\ \mathrm{N} = \mathrm{Molarity} \times \mathrm{n}-\mathrm{factor}\\ \mathrm{n}-\mathrm{factor} =\mathrm{bascity} \mathrm{of} \mathrm{an} \mathrm{acid} = 2 (\mathrm{for} {\mathrm{H}}_2{\mathrm{SO}}_4)\\ \mathrm{molarity}\left(\mathrm{M}\right) = \frac{0.02}{2}=0.01\mathrm{M}\\ \mathrm{M}=\frac{\mathrm{no}.\mathrm{of} \mathrm{moles}\times 1000}{\mathrm{V}\left(\mathrm{ml}\right)}\Rightarrow \mathrm{no}.\mathrm{of} \mathrm{moles}=\frac{0.01\times 100}{1000}=0.001\mathrm{moles}\\ \mathrm{or}\mathrm{Number}\mathrm{of}{\mathrm{H}}_2{\mathrm{SO}}_4\mathrm{molecules}=\text{\hspace{0.17em}\hspace{0.17em}n}\times 6.02\times {10}^{23}=0.001\times 6.02\times {10}^{23}\\ = 6.02\times {10}^{20}\end{array}$