1. Relation between magnetic field and susceptibility

For a magnetic material placed inside a solenoid/toroid, $B = \mu H = \mu_0(1+\chi_m)H$ where $\chi_m$ is the magnetic susceptibility.

If the toroid is empty (air/vacuum), then $B_0 = \mu_0 H$

If the toroid is filled with aluminium, then $B = \mu_0(1+\chi_m)H$

So the increase in magnetic field is $\Delta B = B - B_0 = \mu_0 \chi_m H$

Hence, the fractional increase is $\frac{\Delta B}{B_0} = \chi_m$

2. Given susceptibility

For aluminium, $\chi_m = 2.2 \times 10^{-5}$

Thus, $\frac{\Delta B}{B_0} = 2.2 \times 10^{-5}$

3. Convert to percentage increase

Percentage increase $= \chi_m \times 100$ $= 2.2 \times 10^{-5} \times 100$ $= 2.2 \times 10^{-3}$

Now, $2.2 \times 10^{-3} = 22 \times 10^{-4}$

So the percentage increase is $\frac{22}{10^4}$

Comparing with the given form $\dfrac{x}{10^4}$, $x = 22$

4. Comparison with stored answer

Derived answer is $22$, which matches the stored correct answer.