The volume of 0.05N Na₂CO₃ solution required to neutralise 200ml of 0.02M H₂SO₄ solution is

$\underset{\left(\mathrm{A}\right)}{ {\mathrm{H}}_2{\mathrm{SO}}_4}+\underset{\left(\mathrm{B}\right)}{{\mathrm{Na}}_2{\mathrm{CO}}_3}\to {\mathrm{Na}}_2{\mathrm{SO}}_4+{\mathrm{H}}_2\mathrm{O}+{\mathrm{CO}}_2$

given, V_A=200ml ,  M_A=0.02M

N_A=M$\times$basicity=0.02$\times$2=0.04N

V_B=?  ,  N_B=0.05N

$$\begin{gathered} {\mathrm{V}}_{\mathrm{A}}{\mathrm{N}}_{\mathrm{A}}={\mathrm{V}}_{\mathrm{B}}{\mathrm{N}}_{\mathrm{B}} \\ 200\times 0.04={\mathrm{V}}_{\mathrm{B}}\times 0.05 \\ {\mathrm{V}}_{\mathrm{B}}=\frac{200\times 0.04}{0.05}=160\mathrm{ml} \end{gathered}$$