\(n_{H_2SO_4}=0.3\cdot1.5=0.45\left(mol\right)\)
\(2NaOH+H_2SO_4\rightarrow Na_2SO_4+H_2O\)
\(0.9..............0.45\)
\(m_{NaOH}=0.9\cdot40=36\left(g\right)\)
\(n_{KOH}=0.9\left(mol\right)\)
\(m_{KOH}=0.9\cdot56=50.4\left(g\right)\)
\(m_{dd_{KOH}}=\dfrac{50.4}{5.6\%}=900\left(g\right)\)
\(V_{ddKOH}=\dfrac{900}{1.045}=861.2\left(ml\right)\)