Correct answer is (d) 1.50 × 10^-2 Wb/m^2
To explain I would say: The magnetic field induction at the center of a long solenoid is B = μonI→B ∝ nI n = number of turns per unit length of solenoid. I = current passing through the solenoid.
\(\frac {B1}{B2} = \frac {n1}{n2} [ \frac {I1}{I2} ] \)
\(\frac {B1}{B2} = \frac {500}{300} \, \times \, \frac {I}{(\frac {I}{3})}\)
\(\frac {B1}{B2}\) = 5
B2 = \(\frac {B1}{5}\)
B2 = \(\frac {7.54 \, \times \, 10^{-2}}{5}\)
B2 = 0.01508 = 1.50 × 10^-2Wb/m^2