Given, tension in string $(T)=27 \mathrm{~N}$
$$
\operatorname{Mass}(m)=3 \mathrm{~kg}
$$
Let acceleration of the block be $a$.
$$
\begin{array}{rlrl}
& & 3 g-T & =m a \\
\Rightarrow & & 10 \times 3-27 & =3 a \\
\Rightarrow & & 30-27 & =3 a \\
\Rightarrow & a & =\frac{3}{3}=1 \mathrm{~m} / \mathrm{s}^2
\end{array}
$$
For the body, we have,
$$
\begin{aligned}
& & 27-\mu 20 g & =20 a \\
\Rightarrow & & 27-\mu \times 20 \times 10 & =20 \times 1 \\
\Rightarrow & & 27-200 \mu & =20 \\
\Rightarrow & & 200 \mu & =27-20
\end{aligned}
$$
$\therefore$ Coefficient of kinetic friction,
$$
\mu=\frac{7}{200}=0.035
$$
$\therefore$ Coefficient of kinetic friction is 0.035