You want a field of 6.0 mT inside a long solenoid, with a current of no more than 1.5 A…

Step 1. Turns in each metre. $n = \dfrac{B}{\mu_0 I} = \dfrac{6.0 \times 10^{-3}\ \text{T}}{4\pi \times 10^{-7}\ \text{T m/A} \times 1.5\ \text{A}} = 3183\ \text{m}^{-1}$.

Step 2. Total turns. $N = nL = 3183\ \text{m}^{-1} \times 0.35\ \text{m} = 1114$.

Step 3. The wire. One turn is $\pi d = \pi \times 0.040\ \text{m} = 0.1257$ m. So the length is $1114 \times 0.1257\ \text{m} = 140$ m.