$\begin {array} {1 1} (a)\;-3 & \quad (b)\;-\large\frac{1}{3} \\ (c)\;-\large\frac{1}{2} & \quad (d)\;2 \end {array}$

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At origin the fields due to the two wires adds up and the resultant is doubled.

The field is inversely proportional to the distance from the wire.

By drawing the simple diagram of the situation, one can understand that the field at the two places is in the opposite directions.

Hence, the magnitude of the field at the origin is $2B_0$ and at the point $2a$ is $\large\frac{B_0 + B_0}{3} = \large\frac{2B_0}{3}$

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