Step 1: Set up the force balance.
The electron circles a long charged wire because the wire's electric attraction supplies the centripetal force. So electrostatic force equals centripetal force.
Step 2: Field of a long charged wire.
By Gauss's law, $E = \dfrac{\lambda}{2\pi\varepsilon_0 r}$, so the force on the electron is $F = eE = \dfrac{e\lambda}{2\pi\varepsilon_0 r}$.
Step 3: Write the centripetal requirement.
For circular motion, $\dfrac{e\lambda}{2\pi\varepsilon_0 r} = \dfrac{m v^2}{r}$.
Step 4: Cancel r and solve for lambda.
The $r$ cancels on both sides, giving $\lambda = \dfrac{2\pi\varepsilon_0 m v^2}{e}$.
Step 5: Substitute the numbers.
Using $m = 9.1\times 10^{-31}\,kg$, $v = 4\times 10^7\,ms^{-1}$, $e = 1.6\times 10^{-19}\,C$, and $2\pi\varepsilon_0 = \dfrac{1}{2\times 9\times 10^9} = 5.56\times 10^{-11}$: numerator $m v^2 = 9.1\times 10^{-31}\times 1.6\times 10^{15} = 1.46\times 10^{-15}$. Then $\lambda = \dfrac{5.56\times 10^{-11}\times 1.46\times 10^{-15}}{1.6\times 10^{-19}}$.
Step 6: Compute and conclude.
The numerator is $8.1\times 10^{-26}$, and dividing by $1.6\times 10^{-19}$ gives $\lambda \approx 5\times 10^{-7}\,Cm^{-1}$, which is option (3).
\[ \boxed{\lambda \approx 5\times 10^{-7}\,Cm^{-1}} \]