(A) The magnetic field $B$ produced by a solenoid is given by $B = \mu_0 n I$,where $n$ is the number of turns per unit length and $I$ is the current.
Given $B = 100 \;G = 10^{-2} \;T$.
Using $\mu_0 = 4\pi \times 10^{-7} \;T \;m \;A^{-1}$,we have $n I = \frac{B}{\mu_0} = \frac{10^{-2}}{4\pi \times 10^{-7}} \approx 7958 \;A \;m^{-1}$.
Since the maximum $n = 1000 \;turns \;m^{-1}$ and maximum $I = 15 \;A$,the product $n I$ can be as high as $15000 \;A \;m^{-1}$.
To achieve $n I \approx 7958 \;A \;m^{-1}$,we can choose $n = 800 \;turns \;m^{-1}$ and $I \approx 10 \;A$.
For a uniform field over a length of $10 \;cm$,the solenoid length should be significantly larger,e.g.,$L = 50 \;cm$. The radius should be large enough to accommodate the cross-section,e.g.,$r = 2 \;cm$ (area $\approx 1.25 \times 10^{-3} \;m^2$).
Thus,a solenoid of length $50 \;cm$,radius $2 \;cm$,$400$ turns,and current $10 \;A$ is a suitable design.