$\begin{aligned}
& \text { Let mass of iron }=\mathrm{wg} \
& \Rightarrow \frac{\mathrm{w}}{150 \times 10^3} \times 10^6=12 \
& \Rightarrow \mathrm{w}=150 \times 12 \times 10^{-3}=1.8 \mathrm{gm}
\end{aligned}$
Let mass of FeSO4⋅7H2O=w1gm
$\begin{aligned}
& \Rightarrow \text { Moles of } \mathrm{Fe}=\frac{1.8}{56}=\left(\frac{\mathrm{w}_1}{56+96+7 \times 18}\right) \
& \Rightarrow \mathrm{w}_1=8.935 \mathrm{gm}
\end{aligned}$