When box is dropped from height $h$, its speed when it reaches the ground is

$v=\sqrt{2 g h}$

When block slides down the inclined plane $\theta=45^{\circ}$,

Net downward acceleration of block is

$a=\frac{m g \sin \theta-f}{m}$

where, $f=$ friction force.

$\Rightarrow a=\frac{m g}{m} \sin \theta-\mu m g \cos \theta$

$=g(\sin \theta-\mu \cos \theta)$

$=\frac{g}{\sqrt{2}}(1-\mu)$

${[\because \sin \theta}\left.=\cos \theta=\frac{1}{\sqrt{2}}, \text { when } \theta=45^{\circ}\right]$

Velocity of block when it reaches bottom of inclined plane is

$v^{\prime}=\sqrt{2 a s}$

where, $s=$ slope length of inclined plane.

$\Rightarrow \quad v^{\prime}=\sqrt{2 a h / \sin \theta}$

$=\sqrt{\left(2 \frac{g h}{\sqrt{2}}(1-\mu) \times \sqrt{2}\right)}$

$=\sqrt{2 g h(1-\mu)}$

$=\frac{v}{3} \text { (given) }$

So, $\sqrt{2 g h(1-\mu)}=\frac{1}{3} \sqrt{2 g h}$

$\Rightarrow \quad 1-\mu=\frac{1}{9}$

$\Rightarrow \quad \mu=1-\frac{1}{9}=\frac{8}{9}$