The current required to be passed through a solenoid of 15cm length...

The magnetic intensity is

$H = \dfrac{B}{{\mu}_{0}} - M$

For $M = 0$, it can be written

$H = \left(\dfrac{B}{{\mu}_{0}}\right) = ni$

The data given is

$H = 2 . 4 \times {10}^{3}A{m}^{- 1}l = 15 \times {10}^{- 2}mN = 60$

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

$i = \dfrac{H}{n} = \dfrac{2 . 4 \times {10}^{3} \times 15 \times {10}^{- 2}}{60} = 6A$