The current required to be passed through a solenoid of 15   cm length and 60 turns in order to…

The magnetic intensity is

$H=\frac{B}{{\mu }_0}-M$

For $M=0$, it can be written

$H=\left(\frac{B}{{\mu }_0}\right)=ni$

The data given is 

$$\begin{gathered} H=2.4\times {10}^3 \mathrm{A} {\mathrm{m}}^{-1} \\ l\mathit{=}15\times {10}^{-2} \mathrm{m} \\ N=60 \end{gathered}$$

Using the relation, $n=\frac{N}{l}$ and $H=ni$, the value of the current is,

$i=\frac{H}{n}=\frac{2.4\times {10}^3\times 15\times {10}^{-2}}{60}=6 \mathrm{A}$