1. Interpret 10 ppm of iron in 100 kg wheat

For solids, $10\,\text{ppm}$ means: $10\text{ parts of Fe per }10^6\text{ parts of wheat by mass}$ So, in $100\,\text{kg}$ wheat, required iron mass is: $\frac{10}{10^6}\times 100\,\text{kg} = 10^{-3}\,\text{kg} = 1\,\text{g}$

Thus, we need 1 g of Fe.

2. Find molar mass of ferrous sulphate heptahydrate

Formula: $\mathrm{FeSO_4\cdot 7H_2O}$

Molar mass of $\mathrm{FeSO_4}$: $55.85 + 32.00 + 4(16.00) = 55.85 + 32.00 + 64.00 = 151.85$

Molar mass of $7\mathrm{H_2O}$: $7\times 18 = 126$

Total molar mass: $151.85 + 126 = 277.85\,\text{g mol}^{-1}$

3. Find mass fraction of iron in the salt

Each mole of $\mathrm{FeSO_4\cdot 7H_2O}$ contains $1$ mole Fe, i.e. $55.85$ g Fe.

So, mass fraction of Fe in the salt is: $\frac{55.85}{277.85}$

Hence, mass of salt needed for $1$ g Fe is: $m = \frac{277.85}{55.85}\times 1$

4. Calculate

$m \approx 4.98\,\text{g}$

More precisely, $m = \frac{277.85}{55.85} \approx 4.974\,\text{g}$

5. Final answer

Required amount of $\mathrm{FeSO_4\cdot 7H_2O}$ is: $\boxed{4.97\,\text{g}}$