\(n_{NaOH}=0.4\cdot1.5=0.6\left(mol\right)\)
\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_{2O}\)
\(0.6...........0.3......................0.3\)
\(V_{dd_{H_2SO_4}}=\dfrac{0.3}{1.5}=0.2\left(l\right)\)
\(C_{M_{Na_2SO_4}}=\dfrac{0.3}{0.4+0.2}=0.5\left(M\right)\)