The balanced equation is given as:

\(H_{2}SO_{4} + 2NaOH \to Na_{2}SO_{4} + 2H_{2}O\)

Molarity of \(NaOH = 0.1M\)

Volume of \(NaOH = 20cm^{3}\)

Molarity of \(H_{2}SO_{4} = 0.5M\)

mol NaOH = \(0.1 \times 20cm^{3} = 2molNaOH\)

⇒ \(2mol NaOH \times \frac{1mol H_{2}SO_{4}}{2mol NaOH} = 1 mol H_{2}SO_{4}\)

\(volume of H_{2}SO_{4} = \frac{mole of H_{2}SO_{4}}{molarity of H_{2}SO_{4}}\)

= \(\frac{1 mole}{0.5 M} = 2cm^{3} H_{2}SO_{4}\)