A small square loop of wire of side l is placed inside a large...

Flux linkage for inner loop

$ϕ = {B}_{center} \times {l}^{2}$

$= 4 \times \dfrac{{\mu}_{0}i}{4\pi\dfrac{L}{2}}\left(\sin 45 ° + \sin 45 °\right){l}^{2}$

$⇒ ϕ = 2\sqrt{2}\dfrac{{\mu}_{0}i}{\piL}{l}^{2}$

Therefore, mutual inductance will be

$M = \dfrac{ϕ}{i} = \dfrac{2\sqrt{2}{\mu}_{0}{l}^{2}}{\piL} = 2\sqrt{2}\dfrac{{\mu}_{0}}{\pi}$

$= 2\sqrt{2}\dfrac{4\pi}{\pi} \times {10}^{- 7}$

$= 8\sqrt{2} \times {10}^{- 7}$

$= \sqrt{128} \times {10}^{- 7}H$

Hence, $x = 128$.