The percentage increase in the magnetic field B when the space within a current carrying toroid is filled with aluminium, with a susceptibility  of $2.1\times {10}^{-5}$, is $2.1 \times {10}^{-p}.$ Find $p$.

In the absence of aluminium the magnetic field is  $B_0={\mu }_{0}H$.As the space inside the toroid is filled with aluminium the field becomes

 $B=\mu H={\mu }_0(1+\chi )H$

The increase in the field is  $B-B_0={\mu }_0\chi H$
The percentage increase is  $\frac{B-B_0}{B}\times 100=\frac{{\mu }_0\chi H}{{\mu }_{0}H}\times 100 =$ $\chi \times 100=2.1\times {10}^{-3} \left(\because \chi =2.1\times {10}^{-5}\right)$

By comparison, p=3