The bar magnet coercivity $4\times {10}^3 \mathrm{A} {\mathrm{m}}^{-1}$ i.e., it requires a magnetic intensity $H=4\times {10}^3 \mathrm{A} {\mathrm{m}}^{-1}$to get demagnetised. Let $i$ be the current carried by solenoid having n number of turns per metre length, then by definition $H = ni$. Here$H=4\times 10^3 \mathrm{A} {\mathrm{m}}^{-1}$

$n=\frac{N}{l}=\frac{60}{0.12}=500\mathrm{ }{\mathrm{m}}^{-1}$

$\Rightarrow i=\frac{H}{n}=\frac{4\times 10^3}{500}=8.0\mathrm{ }\mathrm{A}$