$B = mu_0 times (dfrac{N}{L}) times I$

where $mu_0 = 4 pi times 10^{-7}: T.m.A^{-1}$.

According to this relation, a solenoid of length of $20.0:cm$ with $200:loops$ carrying current of $3.25:A$ produces magnetic field of $4.08:T$.

To double the magnetic field with the same solenoid, that is, with same the number of loops and length in the above relation, the value of the current has to be doubled.

The required current to double the magnetic field is $2 times 3.25 = 6.5:A$.

$B = mu_0 times (dfrac{N}{L}) times I$

where $mu_0 = 4 pi times 10^{-7}: T.m.A^{-1}$.

This relation can be rewritten for the current $I$ that produces the magnetic field $B$ within the solenoid that equals Earth’s magnetic field ($5.0 times 10^{-5}:T$) with a solenoid of length $38:cm (= 0.38:m)$ with 430 loops as

$I = dfrac{B}{mu_0} times dfrac{L}{N}$

Substituting the values,

$I = dfrac{5.0 times 10^{-5}}{4 pi times 10^{-7}} times dfrac{0.38}{430} = dfrac{5.0 times 0.38}{4 pi times 430} times 10^2 = 0.0352:A = 35.2:mA$.

$B = mu_0 times (dfrac{N}{L}) times I$

where $mu_0 = 4 pi times 10^{-7}: T.m.A^{-1}$. This relation can be rewritten for the number of loops ($N$) of solenoid of length $75:cm (=0.75:m)$ carrying $8.4:A$ current that produces magnetic field of $1.3:T$ as

$N = dfrac{B}{mu_0} times dfrac{L}{I}$

Substituting the values,

$N = dfrac{1.3}{4 pi times 10^{-7}} times dfrac{0.75}{8.4} = dfrac{1.3 times 0.75}{4 pi times 8.4} times 10^7 = 92367$