How get a dilution at 150g/l of H2SO4 at 96%?
First of all: Never add water to sulphuric acid. The heat of the reaction may lead to an explosion. Use safety glasses, safety gloves, and wear protective clothing.

The densities at 20°C are


Water: $\rho_{water} = 0.9982~\mathrm{g~cm^{-3}}$
15% $\ce{H2SO4}$: $\rho_1 = 1.10~\mathrm{g~cm^{-3}}$
96% $\ce{H2SO4}$: $\rho_2 = 1.84~\mathrm{g~cm^{-3}}$


To get $5000~\mathrm{cm^3}$ 15% $\ce{H2SO4}$ with a mass of

$m = \rho_1 \cdot V = 1.10~\mathrm{g~cm^{-3}} \cdot 5000~\mathrm{cm^3} = 5500~\mathrm{g}$

you need

$m = 0.15 \cdot 5500~\mathrm{g} = 825~\mathrm{g}$ pure $\ce{H2SO4}$

These are contained in

$m = \frac{825~\mathrm{g}}{0.96} = 859.4~\mathrm{g}$ of 96% $\ce{H2SO4}$.

i.e.

$V = \frac{859.4~\mathrm{g}}{1.84~\mathrm{g~cm^{-3}}} = 467~\mathrm{cm^3}$ of 96% $\ce{H2SO4}$.

Finally, the amount of water you need to pour the acid in is

$m = 5500~\mathrm{g} - 859.4~\mathrm{g} = 4640.6~\mathrm{g}$ water, i.e. $4649.0~\mathrm{cm^3}$

In short: gently pour, while stirring, $467~\mathrm{cm^3}$ of 96% $\ce{H2SO4}$ into $4649~\mathrm{cm^3}$ distilled water to get $5000~\mathrm{cm^3}$ 15% $\ce{H2SO4}$. Break the procedure if the temperature gets too high.

Disclaimer: I do not take any responsibility for the correctness of the above calculation.