a,
$n_{NaOH}=\dfrac{200.16\%}{40}=0,8(mol)$
$2NaOH+H_2SO_4\to Na_2SO_4+2H_2O$
$\to n_{Na_2SO_4}=n_{H_2SO_4}=\dfrac{n_{NaOH}}{2}= 0,4(mol)$
$\to a=m_{dd H_2SO_4}=0,4.98:9,8\%=400g$
b,
$m_{dd\text{spu}}=200+400=600g$
$\to C\%_{Na_2SO_4}=\dfrac{0,4.142.100}{600}=9,467\%$