$\sum F_y=0$

$\Rightarrow N =10 g \cos \theta / m _{ A }=10\,kg$

$\sum F_x=m_A \cdot a_A$

$\Rightarrow 10 g \cdot \sin \theta=10 . a _{ A }$

$\Rightarrow a_A=g \sin \theta---$ (i)

$\sum F_y=0$

$N =20 g \cos \theta----$ (ii) $/ m _{ B }=20\,kg$

$\sum F _{ x }= m _{ B } a _{ B }$

$\Rightarrow 20\, g \sin \theta- fk =20 a _{ B }$

$\Rightarrow 20\, g \sin \theta-\mu_{ k } \cdot 20 g \cos \theta=20 a _{ B } g$

$f _{ k }=\mu_{ k } \cdot N$

$=\mu_{ k } \cdot 20\, g \cos \theta$

from $(ii)$

$\Rightarrow \sin \theta-\mu_{ k } \cdot \cos \theta=\frac{a_{ B }}{g}$

$\Rightarrow a _{ B }= g \left(\sin \theta- mu _{ k } \cos \theta\right)---$ (iii)

$(i) \div (iii):$

$\frac{a_A}{a_B}=\frac{g \sin \theta}{g\left(\sin \theta-\mu_k \cos \theta\right)}=\frac{\sin \theta}{\sin \theta\left(1-\mu_{ K } \cdot \cot \theta\right)}$

$\Rightarrow \frac{2}{1}=\frac{1}{1-\mu_{ k }} \Rightarrow\left(1-\mu_{ k }\right) 2=1$

$\theta=45^{\circ}$

$\cot 45^{\circ}=1$

$\Rightarrow 2-2 \mu_{ k }=1 \Rightarrow \mu_{ k }=0.5$