How to calculate the mass of water to add to copper(II) sulfate to form a 4% solution?
A $4~\%$ solution is one that contains $4~\mathrm{g}$ of the solute in $96~\mathrm{g}$ of solvent to give $100~\mathrm{g}$ of solution. Just to get that out of the way.
To calculate the mass of the solute ($m(\ce{CuSO4})$ — not to be confused with $m(\ce{CuSO4 . 5 H2O})$) we can use at least two ways which are mathematically equivalent:

Calculate the amount of $\ce{CuSO4 . 5 H2O}$ — this is equivalent to the amount of $\ce{CuSO4}$. Then calculate the latter’s mass.
$$M(\ce{CuSO4 . 5 H2O}) = 249.72~\mathrm{g\ mol^{-1}}\\[1.2em]
n(\ce{CuSO4 . 5 H2O}) = \frac{m}{M} = \frac{15.0~\mathrm{g}}{249.72~\mathrm{g\ mol^{-1}}} = 6.01 \times 10^{-2}~\mathrm{mol}\\[1.5em]
M(\ce{CuSO4}) = 159.62~\mathrm{g\ mol^{-1}}\\[1.2em]
m(\ce{CuSO4}) = n \times M = 6.01 \times 10^{-2}~\mathrm{mol} \times 159.62~\mathrm{g\ mol^{-1}} = 9.59~\mathrm{g}$$

Use the molar weights of copper(II) sulphate and water to determine the mass percentage of copper(II) sulphate in the salt.
$$\frac{M(\ce{CuSO4})}{M (\ce{CuSO4 . 5 H2O})} = \frac{159.62~\mathrm{g\ mol^{-1}}}{249.72~\mathrm{g\ mol^{-1}}} = 63.9~\%\\[1.5em]
\Longrightarrow m(\ce{CuSO4}) = 15.0~\mathrm{g} \times 63.9~\% = 9.59~\mathrm{g}$$


In any case, we will arrive at a mass of the solute of $9.59~\mathrm{g}$. This is equivalent to the $4~\%$ mentioned at the very top. Therefore:
$$m_\mathrm{tot} = \frac{m_\mathrm{solute}}{0.04} = 240~\mathrm{g}$$
$$m_\mathrm{solvent} = 0.96 \times 240~\mathrm{g} = 230~\mathrm{g}$$
You noted that you already have $5.4~\mathrm{g}$ of water in the copper(II) sulphate hydrate present. Thus, you need to add the difference:
$$m_\mathrm{water} = 230~\mathrm{g} - 5.4~\mathrm{g} = 225~\mathrm{g}$$