$\begin{array}{1 1} 0.007\;mT \\ 7\;mT \\ 0.07\;mT \\ 0.7 \;mT \end{array}$

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The formula for the field inside the solenoid is $B = \mu_0 n I$, where $n = \large\frac{N}{L}$ or the turn density and $I$ is the current.

Given that number of turns, $N = 600$ and the length of the solenoid, $L = 2.7m \rightarrow n = \large\frac{600}{2.7} $$ \approx 222.22 \;m^{-1}$

$\Rightarrow B = 4\pi \times 10^{-7} N/A^{-2} \times 222.22 m^{-1} \times 2.5A \approx 0.7mT$

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